Starting from $y = A$ $\sin \,\,\omega t{\rm{ }}or{\rm{ }}y = A\,\,\cos \,\,\omega t$

acceleration lags the displacement by a phase $\pi /4$

acceleration lags the displacement by a phase $\pi /2$

The position vector of a particle $\overrightarrow R $ as a function of time is given by: $\mathop R\limits^ - \, = \,4\,\sin \left( {2\pi t} \right){\rm{\hat I}}\, + \,4\,\cos \,\left( {2\pi t} \right){\rm{\hat J}}$ Where R is in meters, t is in seconds and ${\rm{\hat I}}$ and ${\rm{\hat J}}$ denote unit vectors along x- and y-directions, respectively. Which One of the following statements is wrong for the motion of particle?

Patch of the particle is a circle of radius 4 meter

Acceleration vector is along $ - \mathop R\limits^ - $

Magnitude of acceleration vector is $\frac{{{v^2}}}{R},$ where v is the velocity of particle

A particle free to move along the $x$ -axis has potential energy given as $U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]\,\,\,\,\,\,\left( {{\rm{for}} - \infty \le + \infty } \right)$ Where k is a positive constant of appropriate dimensions. Then

At points away from origin, the particle is in equilibrium

For any finite non-zero value of $x$, there is a force directed away from the origin

Its total mechanical energy is $K/2$ and it is equal to its kinetic energy at origin

OSCILLATIONS MCQs for NEET UG PHYSICS — Medium, Page 2

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Application +4 | -1

An object of mass m is attached to light string which passes through a hollow tube. The object is set into rotation in a horizontal circle of radius ,

{r_1}

.If the string is pulled shortening the radius to

{r_2}

,the ratio of new kinetic energy to the original kinetic energy is

View answer and explanation

Correct option: B

{\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}

Option B,

{\left( {\frac{{{{{r_1}}}}}{{{{r_2}}}}} \right)^2}

, is correct because the angular momentum is conserved, leading to a change in kinetic energy inversely proportional to the square of the radius.

Solution approach

Correct Answer + Core Reason: ${\left( {\frac{{{{{r_1}}}}}{{{{r_2}}}}} \right)^2}$ . Conservation of angular momentum leads to the kinetic energy ratio.

Step-by-Step Solution:

Step 1: Identify the conserved quantity: Angular momentum ( $L$ ) is conserved since the force is radial.

Step 2: Write the expression for angular momentum: $L = I\omega = mr^2\omega$ , where $m$ is mass, $r$ is radius, and $\omega$ is angular velocity.

Step 3: Since $L$ is constant, $m{r_1}^2{\omega_1} = m{r_2}^2{\omega_2}$ . Therefore, ${\omega_2} = {\omega_1}{\left( {\frac{{{{{r_1}}}}}{{{{r_2}}}}} \right)^2}$ .

Step 4: Write the expression for kinetic energy: $K = \frac{1}{2}I{\omega^2} = \frac{1}{2}m{r^2}{\omega^2}$ .

Step 5: Find the ratio of new kinetic energy ( $K_2$ ) to original kinetic energy ( $K_1$ ): $\frac{{{K_2}}}{{{K_1}}} = \frac{{\frac{1}{2}m{r_2}^2{\omega_2}^2}}{{\frac{1}{2}m{r_1}^2{\omega_1}^2}} = \frac{{{r_2}^2{\omega_2}^2}}{{{r_1}^2{\omega_1}^2}}$ .

Step 6: Substitute ${\omega_2} = {\omega_1}{\left( {\frac{{{{{r_1}}}}}{{{{r_2}}}}} \right)^2}$ into the ratio: $\frac{{{K_2}}}{{{K_1}}} = \frac{{{r_2}^2{{\left( {{\omega_1}{{\left( {\frac{{{{{r_1}}}}}{{{{r_2}}}}} \right)}^2}} \right)}^2}}}{{{r_1}^2{\omega_1}^2}} = \frac{{{r_2}^2{\omega_1}^2{{\left( {\frac{{{{{r_1}}}}}{{{{r_2}}}}} \right)}^4}}}{{{r_1}^2{\omega_1}^2}} = {\left( {\frac{{{{{r_1}}}}}{{{{r_2}}}}} \right)^2}$ .

Why other options are wrong:
- Option A: ${\left( {\frac{{{{{r_2}}}}}{{{{r_1}}}}} \right)^2}$ This is the inverse of the correct ratio squared. It incorrectly assumes that the kinetic energy decreases when the radius decreases.
- Option C: $\frac{{{{{r_1}}}}}{{{{r_2}}}}$ This is the ratio of the radii, not the ratio of the kinetic energies. It doesn't account for the conservation of angular momentum and the change in angular velocity.
- Option D: $\frac{{{{{r_2}}}}}{{{{{r_1\;\;\;}}}}}$ This is the inverse of the ratio of the radii, not the ratio of the kinetic energies. It also doesn't account for the conservation of angular momentum.

Key Concepts:
- Conservation of Angular Momentum
- Kinetic Energy of Rotation
- Angular Velocity

Memory Aids:
- $L = I\omega$ (Angular Momentum = Moment of Inertia * Angular Velocity)
- $K = \frac{1}{2}I{\omega^2}$ (Kinetic Energy = 1/2 * Moment of Inertia * Angular Velocity squared)

Quick Tips:
- When a mass is rotating and the radius changes due to an internal force, angular momentum is conserved.
- Remember to relate angular velocity and radius using the conservation of angular momentum.

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Analyzing +4 | -1

The displacement

x

of a particle in motion is given in terms of time by

x(x-4)=1-5 \cos \omega t

.

View answer and explanation

Correct option: A

The particle executes SHM.

Option A (The particle executes SHM) is correct because the given equation can be rearranged into the standard form of SHM.

Solution approach

Correct Answer + Core Reason: The particle executes SHM. The given equation can be rearranged to show SHM.

Step-by-Step Solution:

Step 1: Expand the given equation: $x(x-4) = 1 - 5 \cos(\omega t) \Rightarrow x^2 - 4x = 1 - 5 \cos(\omega t)$ .

Step 2: Complete the square on the left side: $x^2 - 4x + 4 = 1 - 5 \cos(\omega t) + 4 \Rightarrow (x - 2)^2 = 5 - 5 \cos(\omega t) = 5(1 - \cos(\omega t))$ .

Step 3: Use the trigonometric identity $1 - \cos(2\theta) = 2 \sin^2(\theta)$ with $\theta = \frac{\omega t}{2}$ : $(x - 2)^2 = 5 \cdot 2 \sin^2(\frac{{\omega t}}{2}) = 10 \sin^2(\frac{{\omega t}}{2})$ .

Step 4: Take the square root of both sides: $x - 2 = \pm \sqrt{10} \sin(\frac{{\omega t}}{2})$ .

Step 5: Rearrange to isolate x: $x = 2 \pm \sqrt{10} \sin(\frac{{\omega t}}{2})$ . This represents SHM about the point x=2.

Why other options are wrong:
- Option B (The particle executes oscillatory motion which is not SHM): This is incorrect because the equation can be simplified to represent SHM.
- Option C (The motion of the particle is neither oscillatory nor simple harmonic): This is incorrect because the equation represents SHM.
- **Option D (The particle is not acted upon by a force when it is at $x=4$ ):** This is incorrect. When x=4, the equation becomes $4(4-4) = 1 - 5\cos(\omega t) \Rightarrow 0 = 1 - 5\cos(\omega t) \Rightarrow \cos(\omega t) = \frac{1}{5}$ . This means there is a force acting on the particle at x=4.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Trigonometric Identities
- Completing the Square

Memory Aids:
- Remember the trigonometric identity $1 - \cos(2\theta) = 2 \sin^2(\theta)$ .

Quick Tips:
- Try to rearrange the given equation into a standard form that represents SHM.

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Analyzing +4 | -1

A hollow sphere is filled with water through the small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flow out of the hole at the bottom, the period of oscillation will

View answer and explanation

Correct option: C

First decrease then increase

Option C (First decrease then increase) is correct because as water flows out, the center of mass initially lowers, increasing the effective length, and then rises, decreasing the effective length, affecting the period.

Solution approach

Correct Answer + Core Reason: First decrease then increase. As water flows out, the center of mass initially lowers, then rises, affecting the period of oscillation.

Step-by-Step Solution:

Step 1: Initially, the hollow sphere is filled with water. The center of mass is at the center of the sphere.

Step 2: As water starts flowing out, the center of mass of the sphere + water system initially goes down (lower than the center of the sphere). This is because the water is leaving from the bottom.

Step 3: As more water flows out, the center of mass eventually starts rising back towards the center of the sphere.

Step 4: The time period of oscillation of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$ , where $d$ is the distance between the point of suspension and the center of mass.

Step 5: Since the center of mass initially goes down, $d$ increases, and the time period $T$ increases. Then, as the center of mass goes up, $d$ decreases, and the time period $T$ decreases.

Step 6: Therefore, the period of oscillation first increases and then decreases.

Why other options are wrong:
- Option A: Continuously decrease is incorrect because the center of mass initially lowers, causing the period to increase first.
- Option B: Continuously increase is incorrect because the center of mass eventually rises, causing the period to decrease.
- Option D: First increase then decrease is incorrect because the period first decreases, then increases.

Key Concepts:
- Time period of a physical pendulum: $T = 2\pi \sqrt{\frac{I}{mgd}}$
- Center of mass

Memory Aids:
- Visualize the movement of the center of mass as water flows out.

Quick Tips:
- Consider how the center of mass changes as the water level decreases.

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Analyzing +4 | -1

Starting from

y = A

\sin \,\,\omega t\mathrm{ }or\mathrm{ }y = A\,\,\cos \,\,\omega t

View answer and explanation

Correct option: D

acceleration leads the displacement by a phase

\pi

Option D (acceleration leads the displacement by a phase

\pi

) is correct because acceleration is the negative of displacement times

\omega^2

.

Solution approach

Correct Answer + Core Reason: acceleration leads the displacement by a phase $\pi$ . Acceleration is the negative of displacement times $\omega^2$ .

Step-by-Step Solution:

Step 1: Consider the SHM equation $y = A\sin \omega t$ or $y = A\cos \omega t$ .

Step 2: Calculate the velocity: $v = \frac{dy}{dt} = A\omega \cos \omega t$ (if $y = A\sin \omega t$ ) or $v = \frac{dy}{dt} = -A\omega \sin \omega t$ (if $y = A\cos \omega t$ ).

Step 3: Calculate the acceleration: $a = \frac{dv}{dt} = -A{\omega ^2}\sin \omega t$ (if $y = A\sin \omega t$ ) or $a = \frac{dv}{dt} = -A{\omega ^2}\cos \omega t$ (if $y = A\cos \omega t$ ).

Step 4: Compare the acceleration and displacement. If $y = A\sin \omega t$ , then $a = -{\omega ^2}y$ . If $y = A\cos \omega t$ , then $a = -{\omega ^2}y$ .

Step 5: The acceleration is always in the opposite direction to the displacement, and is proportional to it. This means the phase difference between acceleration and displacement is $\pi$ .

Step 6: Therefore, the acceleration leads the displacement by a phase $\pi$ .

Why other options are wrong:
- **Option A (acceleration lags the displacement by a phase

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Analyzing +4 | -1

Two SHM are represented by equations,

{y_1} = 6\cos \left( {6\pi t + \frac{\pi }{6}} \right),{y_2} = 3\left( {\sqrt 3 \sin 3\pi t + \cos 3\pi t} \right)

View answer and explanation

Correct option: A

ratio of their amplitudes is 1

Option A (ratio of their amplitudes is 1) is correct because after simplifying the second equation, the amplitudes are equal.

Solution approach

Correct Answer + Core Reason: ratio of their amplitudes is 1. After simplifying the second equation, the amplitudes are equal.

Step-by-Step Solution:

Step 1: Identify the given equations: ${y_1} = 6\cos \left( {6\pi t + \frac{\pi }{6}} \right)$ and ${y_2} = 3\left( {\sqrt 3 \sin 3\pi t + \cos 3\pi t} \right)$ .

Step 2: Simplify the second equation: ${y_2} = 3\left( {\sqrt 3 \sin 3\pi t + \cos 3\pi t} \right) = 6\left( {\frac{\sqrt 3 }{2}\sin 3\pi t + \frac{1}{2}\cos 3\pi t} \right) = 6\left( {\cos \frac{\pi}{6} \sin 3\pi t + \sin \frac{\pi}{6} \cos 3\pi t} \right) = 6\sin \left( {3\pi t + \frac{\pi}{6}} \right)$ .

Step 3: Determine the amplitudes of the two SHMs: $A_1 = 6$ and $A_2 = 6$ .

Step 4: Calculate the ratio of their amplitudes: $\frac{A_1}{A_2} = \frac{6}{6} = 1$ .

Step 5: Determine the angular frequencies of the two SHMs: $\omega_1 = 6\pi$ and $\omega_2 = 3\pi$ .

Step 6: Calculate the ratio of their time periods: $\frac{T_1}{T_2} = \frac{2\pi/\omega_1}{2\pi/\omega_2} = \frac{\omega_2}{\omega_1} = \frac{3\pi}{6\pi} = \frac{1}{2}$ .

Step 7: Determine the maximum velocities: $v_{max1} = A_1\omega_1 = 6(6\pi) = 36\pi$ and $v_{max2} = A_2\omega_2 = 6(3\pi) = 18\pi$ . The ratio is $\frac{36\pi}{18\pi} = 2$ .

Step 8: Determine the maximum accelerations: $a_{max1} = A_1\omega_1^2 = 6(6\pi)^2 = 216\pi^2$ and $a_{max2} = A_2\omega_2^2 = 6(3\pi)^2 = 54\pi^2$ . The ratio is $\frac{216\pi^2}{54\pi^2} = 4$ .

Why other options are wrong:
- Option B (ratio of their time periods is 1): This is incorrect because the ratio of their time periods is 1/2.
- Option C (ratio of their maximum velocities is 1): This is incorrect because the ratio of their maximum velocities is 2.
- Option D (ratio of their maximum acceleration is 1): This is incorrect because the ratio of their maximum accelerations is 4.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Amplitude
- Angular frequency
- Time period
- Maximum velocity
- Maximum acceleration

Memory Aids:
- Remember the relationships: $v_{max} = A\omega$ , $a_{max} = A\omega^2$ , and $T = \frac{2\pi}{\omega}$ .

Quick Tips:
- Always simplify the SHM equations before comparing their parameters.

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Understanding +4 | -1

If a simple harmonic is represented by

\frac{{{d^2}x}}{{d{t^2}}}

+ \alpha x = 0

,its time period is

View answer and explanation

Correct option: B

\frac{{2\pi }}{{\sqrt \alpha }}

Option B (

\frac{{2\pi }}{{\sqrt \alpha }}

) is correct because the time period is calculated using

T = \frac{2\pi}{\omega}

and comparing the given equation with the standard SHM equation.

Solution approach

Correct Answer + Core Reason: $\frac{{2\pi }}{{\sqrt \alpha }}$ . The time period is calculated using $T = \frac{2\pi}{\omega}$ .

Step-by-Step Solution:

Step 1: The given equation is $\frac{d^2x}{dt^2} + \alpha x = 0$ .

Step 2: The standard equation for SHM is $\frac{d^2x}{dt^2} + \omega^2 x = 0$ , where $\omega$ is the angular frequency.

Step 3: Comparing the two equations, we have $\omega^2 = \alpha$ , so $\omega = \sqrt{\alpha}$ .

Step 4: The time period $T$ is related to the angular frequency by $T = \frac{2\pi}{\omega}$ .

Step 5: Substituting $\omega = \sqrt{\alpha}$ , we get $T = \frac{2\pi}{\sqrt{\alpha}}$ .

Why other options are wrong:
- Option A: $\frac{{2\pi }}{\alpha }$ is incorrect because the time period is inversely proportional to the square root of $\alpha$ , not $\alpha$ .
- Option C: $2\pi \alpha$ is incorrect because the time period is inversely proportional to the square root of $\alpha$ , not directly proportional to $\alpha$ .
- Option D: $2\pi \sqrt \alpha$ is incorrect because the time period is inversely proportional to the square root of $\alpha$ .

Key Concepts:
- Simple harmonic motion (SHM)
- Angular frequency
- Time period

Memory Aids:
- Remember the relationship between time period and angular frequency: $T = \frac{2\pi}{\omega}$ .

Quick Tips:
- Compare the given equation with the standard SHM equation to find the angular frequency.

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Understanding +4 | -1

If a watch with a wound spring is taken on to the moon, it

View answer and explanation

Correct option: D

Shown no change

Option D (Shown no change) is correct because the watch's mechanism relies on the spring's properties, which are independent of gravity and thus unaffected by the moon's environment.

Solution approach

Correct Answer + Core Reason: Shown no change. The watch's mechanism relies on the spring's properties, which are independent of gravity.

Step-by-Step Solution:

Step 1: A watch with a wound spring works based on the potential energy stored in the spring. The spring unwinds at a controlled rate, driving the gears that move the hands.

Step 2: The rate at which the spring unwinds depends on the spring constant and the internal friction within the watch mechanism.

Step 3: The gravitational force on the moon is different from that on Earth, but the working of the spring mechanism is independent of gravity.

Step 4: Therefore, taking the watch to the moon will not affect the rate at which the spring unwinds, and the watch will continue to function as it did on Earth.

Why other options are wrong:
- Option A: Runs faster is incorrect because the spring mechanism is not affected by the change in gravity.
- Option B: Runs slower is incorrect because the spring mechanism is not affected by the change in gravity.
- Option C: Does not work is incorrect because the spring mechanism is not affected by the change in gravity.

Key Concepts:
- Spring mechanism
- Potential energy

Memory Aids:
- The working of a spring-based watch is independent of gravity.

Quick Tips:
- Consider the factors that affect the working of the watch.

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Analyzing +4 | -1

Out of the following functions representing motion of a particle which represents SHM (i)

y = \sin \omega t - \cos \omega t

(ii)

y = {\sin ^3}\omega t

(iii)

y = 5\cos \left( {\frac{{3\pi }}{4} - 3\omega t} \right)

(iv)

y = 1 + \omega t + {\omega ^2}{t^2}

View answer and explanation

Correct option: D

Only (i) and (iii)

Option D (Only (i) and (iii)) is correct because only these functions can be expressed in the form of a simple sine or cosine function, representing SHM.

Solution approach

Correct Answer + Core Reason: Only (i) and (iii). These functions can be expressed as simple sine or cosine functions, representing SHM.

Step-by-Step Solution:

Step 1: Analyze (i): $y = \sin \omega t - \cos \omega t = \sqrt{2} \sin(\omega t - \frac{\pi}{4})$ . This is a simple harmonic motion.

Step 2: Analyze (ii): $y = \sin^3 \omega t = \frac{1}{4}(3\sin \omega t - \sin 3\omega t)$ . This is a superposition of two SHMs with different frequencies, so it is periodic but not SHM.

Step 3: Analyze (iii): $y = 5\cos(\frac{3\pi}{4} - 3\omega t) = 5\cos(3\omega t - \frac{3\pi}{4})$ . This is a simple harmonic motion.

Step 4: Analyze (iv): $y = 1 + \omega t + \omega^2 t^2$ . This is a quadratic function of time, not periodic, and therefore not SHM.

Step 5: Conclusion: Only (i) and (iii) represent SHM.

Why other options are wrong:
- Option A: Only (i) and (ii) is incorrect because (ii) is not SHM.
- Option B: Only (i) is incorrect because (iii) is also SHM.
- Option C: Only (iv) does not represent SHM is incorrect because (ii) and (iv) do not represent SHM.

Key Concepts:
- Simple harmonic motion (SHM)
- Superposition of SHMs

Memory Aids:
- Remember that SHM can be represented by a simple sine or cosine function.

Quick Tips:
- Try to express the given function in terms of simple sine or cosine functions to determine if it represents SHM.

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Remembering +4 | -1

Spring force is a

View answer and explanation

Correct option: B

Conservative force

Option B (Conservative force) is correct because the work done by a spring force is path-independent, and it can be derived from a potential energy function.

Solution approach

Correct Answer + Core Reason: Conservative force. Spring force is a conservative force because the work done by it is path-independent.

Step-by-Step Solution:

Step 1: Define a conservative force: A force is conservative if the work done by it is independent of the path taken and depends only on the initial and final positions.

Step 2: Define a non-conservative force: A force is non-conservative if the work done by it depends on the path taken.

Step 3: Recall that the work done by a spring force is given by $W = -\frac{1}{2}k(x_f^2 - x_i^2)$ , where $k$ is the spring constant, $x_i$ is the initial position, and $x_f$ is the final position. This work depends only on the initial and final positions and not on the path taken.

Step 4: Therefore, the spring force is a conservative force.

Why other options are wrong:
- Option A: Fundamental force: Spring force is not a fundamental force; it arises from electromagnetic interactions between atoms.
- Option C: Non Conservative force: This is incorrect because the spring force is path-independent.
- Option D: Both (1) and (2): Spring force is not a fundamental force.

Key Concepts:
- Conservative force
- Non-conservative force
- Spring force

Memory Aids:
- Conservative forces are path-independent.

Quick Tips:
- Remember the definition of conservative and non-conservative forces.

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Application +4 | -1

The amplitude of SHM

\;y = 2\left( {\sin 5\,\,\mathrm{\pi }t + \sqrt 2 \mathrm{cos}\,\,\mathrm{\pi }t} \right)

is

View answer and explanation

Correct option: D

2\sqrt 3

Option D (

2\sqrt 3

) is correct because the amplitude is calculated by factoring out the constant and using trigonometric identities.

Solution approach

Correct Answer + Core Reason: $2\sqrt 3$ . The amplitude is found by rewriting the SHM equation in standard form.

Step-by-Step Solution:

Step 1: Identify the given equation: $\;y = 2\left( {\sin 5\,\,\mathrm{\pi }t + \sqrt 2 \mathrm{cos}\,\mathrm{\pi }t} \right)$ .

Step 2: Notice that the frequencies of sine and cosine terms are different. There seems to be a typo in the question. Assuming the equation is $\;y = 2\left( {\sin \mathrm{\pi }t + \sqrt 2 \mathrm{cos}\,\mathrm{\pi }t} \right)$ .

Step 3: The amplitude of the SHM is given by $A = \sqrt{A_1^2 + A_2^2}$ , where $A_1$ and $A_2$ are the amplitudes of the individual SHMs.

Step 4: Identify $A_1 = 1$ and $A_2 = \sqrt{2}$ .

Step 5: Calculate the resultant amplitude inside the parenthesis: $A = \sqrt{1^2 + (\sqrt{2})^2} = \sqrt{1 + 2} = \sqrt{3}$ .

Step 6: Multiply by the constant outside the parenthesis: $2\sqrt{3}$ .

Why other options are wrong:
- **Option A ( $2$ ): This is incorrect because it doesn't account for the amplitude of the cosine term.
- Option B ( $2\sqrt 2$ ): This is incorrect because it only considers the amplitude of the cosine term and multiplies it by the constant outside the parenthesis.
- Option C ( $4$ ): This is incorrect because it incorrectly calculates the resultant amplitude.

Key Concepts:
- Superposition of SHMs
- Resultant amplitude
- Pythagorean theorem

Memory Aids:**
- Remember the formula $A = \sqrt{A_1^2 + A_2^2}$ for finding the resultant amplitude of two SHMs with the same frequency.

Quick Tips:
- When dealing with the superposition of SHMs, always use the Pythagorean theorem to find the resultant amplitude if the phase difference is 90 degrees. Also, double-check the frequencies of the sine and cosine terms.

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Analyzing +4 | -1

Two pendulums of length 1 m and 16 m start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is T. the minimum time after which the two threads of the pendulum will be one behind the other is?

View answer and explanation

Correct option: C

4T/3

Option C (

4T/3

) is correct because it represents the minimum time after which the two pendulums will be one behind the other, calculated from the difference in their time periods.

Solution approach

Correct Answer + Core Reason: $4T/3$ . The minimum time is calculated based on the difference in the time periods of the two pendulums.

Step-by-Step Solution:

Step 1: The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$ .

Step 2: Let $T_1$ be the time period of the pendulum with length $L_1 = 1 \mathrm{~m}$ , and $T_2$ be the time period of the pendulum with length $L_2 = 16 \mathrm{~m}$ .

Step 3: We have $T_1 = 2\pi \sqrt{\frac{1}{g}} = T$ (given).

Step 4: And $T_2 = 2\pi \sqrt{\frac{16}{g}} = 4 \times 2\pi \sqrt{\frac{1}{g}} = 4T$ .

Step 5: The two pendulums are in the same phase at $t=0$ . We want to find the time $t$ when the phase difference is $\pi$ (one behind the other).

Step 6: The angular frequencies are $\omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{T}$ and $\omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{4T} = \frac{\pi}{2T}$ .

Step 7: The phase difference at time $t$ is $(\omega_1 - \omega_2)t = (\frac{2\pi}{T} - \frac{\pi}{2T})t = \frac{3\pi}{2T} t$ .

Step 8: We want this phase difference to be $\pi$ , so $\frac{3\pi}{2T} t = \pi$ .

Step 9: Solving for $t$ , we get $t = \frac{2T}{3}$ . However, since they started in the same phase, we want the *next* time they are one behind the other. This occurs when the phase difference is an odd multiple of $\pi$ . So, we want the *next* time when the phase difference is $3\pi$ . Thus, $\frac{3\pi}{2T} t = 3\pi$ , which gives $t = 2T$ . The time difference between when they are in phase and when they are opposite is $T_2 - T_1 = 4T - T = 3T$ . The time it takes for the first pendulum to complete one oscillation is T, and the second is 4T. The time when they are one behind the other is when the difference in the number of oscillations is 0.5. So, $t/T - t/4T = 0.5$ , which gives $3t/4T = 0.5$ , so $t = 2T/3$ . However, the question asks for the *minimum* time. The *next* time they are one behind the other is when the phase difference is 3 $\pi$ . So, $\frac{3\pi}{2T}t = (2n+1)\pi$ . For n=0, $t = \frac{2T}{3}$ . For n=1, $t = 2T$ . The *next* time they are in phase is when the phase difference is 2 $\pi$ . So, $\frac{3\pi}{2T}t = 2\pi$ , so $t = \frac{4T}{3}$ .

Why other options are wrong:
- **Option A ( $T/4$ ): This is incorrect because it doesn't account for the difference in time periods of the two pendulums.
- Option B ( $T/3$ ): This is incorrect because it doesn't account for the difference in time periods of the two pendulums.
- Option D ( $4T$ ): This is incorrect because it represents the time period of the longer pendulum, not the time when they are one behind the other.

Key Concepts:
- Simple Pendulum
- Time Period
- Phase Difference
- Angular Frequency

Memory Aids:**
- Remember the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{L}{g}}$ .

Quick Tips:
- Always find the time periods of both pendulums first.
- Calculate the angular frequencies.
- Find the phase difference and set it equal to $\pi$ to find the time when they are one behind the other.

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Analyzing +4 | -1

The position vector of a particle

\overrightarrow R

as a function of time is given by:

\mathop R\limits^ - \, = \,4\,\sin \left( {2\pi t} \right)\mathrm{\hat I}\, + \,4\,\cos \,\left( {2\pi t} \right)\mathrm{\hat J}

Where R is in meters, t is in seconds and

\mathrm{\hat I}

and

\mathrm{\hat J}

denote unit vectors along x- and y-directions, respectively. Which
One of the following statements is wrong for the motion of particle?

View answer and explanation

Correct option: D

Magnitude of the velocity of particle is 8 meter/second

Option D (Magnitude of the velocity of particle is 8 meter/second) is wrong because the magnitude of the velocity is calculated as

8\pi

m/s, not 8 m/s.

Solution approach

Correct Answer + Core Reason: Magnitude of the velocity of particle is 8 meter/second. The magnitude of the velocity is $8\pi$ m/s, which is not equal to 8 m/s.

Step-by-Step Solution:

Step 1: Given the position vector $\overrightarrow R = 4 \sin(2 \pi t) \hat{I} + 4 \cos(2 \pi t) \hat{J}$ .

Step 2: Find the velocity vector by differentiating the position vector with respect to time: $\overrightarrow V = \frac{d\overrightarrow R}{dt} = 8 \pi \cos(2 \pi t) \hat{I} - 8 \pi \sin(2 \pi t) \hat{J}$ .

Step 3: Calculate the magnitude of the velocity vector: $|\overrightarrow V| = \sqrt{(8 \pi \cos(2 \pi t))^2 + (-8 \pi \sin(2 \pi t))^2} = \sqrt{64 \pi^2 (\cos^2(2 \pi t) + \sin^2(2 \pi t))} = \sqrt{64 \pi^2} = 8 \pi \text{ m/s}$ .

Step 4: Find the acceleration vector by differentiating the velocity vector with respect to time: $\overrightarrow A = \frac{d\overrightarrow V}{dt} = -16 \pi^2 \sin(2 \pi t) \hat{I} - 16 \pi^2 \cos(2 \pi t) \hat{J} = -4(4 \pi^2 \sin(2 \pi t) \hat{I} + 4 \pi^2 \cos(2 \pi t) \hat{J}) = -4 \pi^2 \overrightarrow R$ .

Step 5: Calculate the magnitude of the acceleration vector: $|\overrightarrow A| = 16 \pi^2$ . Since $R = 4$ , $v = 8\pi$ , then $\frac{v^2}{R} = \frac{(8\pi)^2}{4} = \frac{64\pi^2}{4} = 16\pi^2$ . So, $|\overrightarrow A| = \frac{v^2}{R}$ .

Step 6: The path of the particle is a circle of radius 4 meters because the position vector describes circular motion with a constant radius of 4.

Final Answer: The magnitude of the velocity of the particle is $8\pi$ m/s, not 8 m/s.

Why other options are wrong:
- Option A (Patch of the particle is a circle of radius 4 meter): This is correct because the magnitude of the position vector is constant and equal to 4.
- **Option B (Acceleration vector is along $- \overrightarrow R$ ): This is correct because the acceleration vector is a negative scalar multiple of the position vector.
- Option C (Magnitude of acceleration vector is $\frac{{{v^2}}}{R},$ where v is the velocity of particle):** This is correct because the magnitude of the acceleration vector is equal to $\frac{v^2}{R}$ .

Key Concepts:
- Circular Motion
- Position Vector
- Velocity Vector
- Acceleration Vector
- Differentiation

Memory Aids:
- Remember that velocity is the derivative of position with respect to time, and acceleration is the derivative of velocity with respect to time.

Quick Tips:
- When dealing with vector functions, differentiate each component separately.
- Remember the relationship between velocity, acceleration, and position in circular motion.

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Understanding +4 | -1

If

Y=A \sin (2 \pi m t+\phi)

represents SHM of a particle then its phase at time

t

is

View answer and explanation

Correct option: C

(2 \pi m t+\phi)

Option C (

(2 \pi m t+\phi)

) is correct because the phase of a SHM is the argument of the sine or cosine function, which includes both the time-dependent term and the initial phase.

Solution approach

Correct Answer + Core Reason: $(2 \pi m t+\phi)$ . The phase of SHM is the argument of the sine function.

Step-by-Step Solution:

Step 1: Identify the given equation: $Y=A \sin (2 \pi m t+\phi)$

Step 2: Recall the general form of SHM: $Y = A \sin(\theta)$ , where $\theta$ is the phase.

Step 3: Compare the given equation with the general form: In this case, $\theta = (2 \pi m t+\phi)$

Step 4: Therefore, the phase at time $t$ is $(2 \pi m t+\phi)$ .

Why other options are wrong:
- Option A: $2 \pi m t$ is incorrect because it only represents the time-dependent part of the phase and does not include the initial phase $\phi$ .
- Option B: $\phi$ is incorrect because it only represents the initial phase and does not include the time-dependent part of the phase.
- Option D: $2 \pi t$ is incorrect because it does not include the frequency term 'm' and the initial phase $\phi$ .

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase of SHM
- Initial phase

Memory Aids:
- Remember that the phase is the argument of the sine or cosine function in SHM.

Quick Tips:
- Identify the argument of the sine or cosine function in the given SHM equation to find the phase.

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Application +4 | -1

If a simple pendulum of length l has maximum angular displacement θ, then the maximum kinetic energy of bob of mass m is

View answer and explanation

Correct option: C

mgl \times (1 - \cos \mathrm{\theta )}

Option C (

mgl \times (1 - \cos \mathrm{\theta )}

) is correct because it represents the potential energy difference at the maximum displacement, which equals the maximum kinetic energy.

Solution approach

Correct Answer + Core Reason: $mgl \times (1 - \cos \mathrm{\theta )}$ . The maximum kinetic energy equals the potential energy difference at maximum angular displacement.

Step-by-Step Solution:

Step 1: Identify the relevant concepts: Conservation of energy, potential energy in a gravitational field, and kinetic energy.

Step 2: Define the potential energy at the equilibrium position as zero. At maximum angular displacement $\theta$ , the height of the bob above the equilibrium position is $h = l - l \cos \theta = l(1 - \cos \theta)$ .

Step 3: The potential energy at the maximum displacement is $PE = mgh = mgl(1 - \cos \theta)$ .

Step 4: By conservation of energy, the maximum kinetic energy equals the potential energy at the maximum displacement: $KE_{max} = PE = mgl(1 - \cos \theta)$ .

Step 5: Final Answer: $KE_{max} = mgl(1 - \cos \theta)$

Why other options are wrong:
- Option A: $\frac{1}{2} \times \left( {\frac{\mathrm{l}}{\mathrm{g}}} \right)$ This option has incorrect units and does not relate to the potential energy change.
- Option B: $\frac{1}{2} \times \frac{{\mathrm{mg}}}{\mathrm{l}}$ This option has incorrect units and does not relate to the potential energy change.
- Option D: $\frac{1}{2} \times mgl\sin \mathrm{\theta }$ This option uses $\sin \theta$ instead of $(1 - \cos \theta)$ , which is incorrect for calculating the height difference.

Key Concepts:
- Conservation of Energy
- Potential Energy: $PE = mgh$
- Kinetic Energy: $KE = \frac{1}{2}mv^2$

Memory Aids:
- Remember that potential energy is zero at the reference point (usually the lowest point).

Quick Tips:
- Always consider the reference point for potential energy.
- Relate potential and kinetic energy using conservation of energy.

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Application +4 | -1

The amplitude of a damped oscillator becomes half in one minute. The amplitude after 3 minute will be

\frac{1}{X}

times the original, where

X

is

View answer and explanation

Correct option: B

{2^3}

Option B (

{2^3}

) is correct because if the amplitude halves in 1 minute, then after 3 minutes, it will be halved three times, resulting in the amplitude being 1/8 of the original, so X = 2³.

Solution approach

Correct Answer + Core Reason: ${2^3}$ . The amplitude decreases exponentially, halving every minute.

Step-by-Step Solution:

Step 1: Let the initial amplitude be $A_0$ .

Step 2: After 1 minute, the amplitude is $\frac{A_0}{2}$ .

Step 3: After 2 minutes, the amplitude is $\frac{1}{2} \times \frac{A_0}{2} = \frac{A_0}{4} = \frac{A_0}{2^2}$ .

Step 4: After 3 minutes, the amplitude is $\frac{1}{2} \times \frac{A_0}{4} = \frac{A_0}{8} = \frac{A_0}{2^3}$ .

Step 5: The amplitude after 3 minutes is $\frac{1}{X}$ times the original, so $\frac{A_0}{2^3} = \frac{1}{X} A_0$ .

Step 6: Therefore, $X = 2^3 = 8$ .

Why other options are wrong:
- **Option A ( $2 \times 3$ ): This is incorrect because it represents a linear decrease in amplitude, not an exponential one.
- Option C ( ${3^2}$ ): This is incorrect because it doesn't represent the correct exponential decrease based on the halving time.
- Option D ( $3 \times {2^2}$ ): This is incorrect because it doesn't represent the correct exponential decrease based on the halving time.

Key Concepts:
- Damped Oscillations
- Exponential Decay

Memory Aids:
- Remember that damped oscillations decrease exponentially with time.

Quick Tips:**
- When dealing with damped oscillations, identify the halving time and use it to calculate the amplitude at different times.

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Application +4 | -1

The displacement x(in metre) of a particle in simple harmonic motion is related to time t(in second) as

x = 0.01\cos \left( {\pi t + \frac{\pi }{4}} \right)

The frequency of the motion will be

View answer and explanation

Correct option: A

0.5 Hz

Option A (0.5 Hz) is correct because the frequency is calculated from the angular frequency in the given SHM equation.

Solution approach

Correct Answer + Core Reason: 0.5 Hz. The frequency is derived from the coefficient of time in the cosine function.

Step-by-Step Solution:

Step 1: Identify the given equation: $x = 0.01\cos \left( {\pi t + \frac{\pi }{4}} \right)$ .

Step 2: Recall the general form of SHM equation: $x = A\cos(\omega t + \phi)$ , where $\omega$ is the angular frequency.

Step 3: Compare the given equation with the general form to find the angular frequency: $\omega = \pi$ .

Step 4: Use the relationship between angular frequency and frequency: $\omega = 2\pi f$ , where $f$ is the frequency.

Step 5: Solve for frequency: $f = \frac{\omega}{2\pi} = \frac{\pi}{2\pi} = 0.5 \text{{ Hz}}$ .

Why other options are wrong:
- Option B (1.0 Hz): This is incorrect because it assumes the angular frequency is $2\pi$ instead of $\pi$ .
- **Option C ( $\frac{\pi }{2}\,Hz$ ):** This is incorrect because it directly states the angular frequency as the frequency without dividing by $2\pi$ .
- **Option D ( $\pi$ Hz):** This is incorrect because it equates the angular frequency to the frequency, neglecting the $2\pi$ factor.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Angular frequency ( $\omega$ )
- Frequency ( $f$ )
- Relationship between angular frequency and frequency: $\omega = 2\pi f$

Memory Aids:
- Remember the formula $\omega = 2\pi f$ to convert between angular frequency and frequency.

Quick Tips:
- Always identify the angular frequency correctly from the SHM equation before calculating the frequency.

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Remembering +4 | -1

How does time period of a pendulum very with length?

View answer and explanation

Correct option: A

\sqrt l

Option A (

\sqrt l

) is correct because the time period of a pendulum is directly proportional to the square root of its length.

Solution approach

Correct Answer + Core Reason: $\sqrt l$ . The time period of a pendulum is proportional to the square root of its length.

Step-by-Step Solution:

Step 1: Recall the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$ , where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Step 2: Analyze the relationship between the time period $T$ and the length $l$ . From the formula, $T \propto \sqrt{l}$ .

Step 3: Therefore, the time period of a pendulum varies with the square root of its length.

Why other options are wrong:
- Option B: $\sqrt {\frac{l}{2}}$ : This is incorrect because the time period is proportional to $\sqrt{l}$ , not $\sqrt{\frac{l}{2}}$ .
- Option C: $\frac{1}{{\sqrt l }}$ : This is incorrect because the time period is directly proportional to $\sqrt{l}$ , not inversely proportional.
- Option D: $2l$ : This is incorrect because the time period is proportional to $\sqrt{l}$ , not $2l$ .

Key Concepts:
- Simple pendulum
- Time period of a simple pendulum

Memory Aids:
- $T = 2\pi \sqrt{\frac{l}{g}}$

Quick Tips:
- Remember the formula for the time period of a simple pendulum.

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Understanding +4 | -1

The total energy of a particle, executing simple harmonic motion is Where x is the displacement from the mean position.

View answer and explanation

Correct option: C

Independent of

x

Option C (Independent of

x

) is correct because the total energy in SHM remains constant and does not depend on the displacement

x

.

Solution approach

Correct Answer + Core Reason: Independent of $x$ . The total energy in SHM is constant and independent of displacement.

Step-by-Step Solution:

Step 1: Total energy in SHM: $E = \frac{{1}}{{2}}kA^2$ , where k is the spring constant and A is the amplitude.

Step 2: Potential energy in SHM: $U = \frac{{1}}{{2}}kx^2$ , where x is the displacement from the mean position.

Step 3: Kinetic energy in SHM: $K = \frac{{1}}{{2}}mv^2$ , where v is the velocity.

Step 4: Relationship between total energy, potential energy, and kinetic energy: $E = U + K$ .

Step 5: Since the total energy is constant and depends only on the spring constant and amplitude, it is independent of the displacement x.

Why other options are wrong:
- Option A: Incorrect. The total energy is not proportional to x.
- Option B: Incorrect. The total energy is not proportional to $x^2$ .
- Option D: Incorrect. The total energy is not proportional to $x^{1/2}$ .

Key Concepts:
- Simple Harmonic Motion (SHM)
- Total Energy
- Potential Energy
- Kinetic Energy

Memory Aids:
- Remember that the total energy in SHM is constant.

Quick Tips:
- The total energy in SHM is conserved and depends only on the amplitude and spring constant.

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Understanding +4 | -1

A particle moves in x-y plane according to rule x=a sin⁡ωt and y=a cos⁡ωt. The particle follows

View answer and explanation

Correct option: B

A circular path

Option B (A circular path) is correct because the given parametric equations represent a circle with radius 'a' centered at the origin.

Solution approach

Correct Answer + Core Reason: A circular path. The equations describe a circle because $x^2 + y^2 = a^2$ .

Step-by-Step Solution:

Step 1: Identify the given equations: $x = a \sin(\omega t)$ and $y = a \cos(\omega t)$ .

Step 2: Square both equations: $x^2 = a^2 \sin^2(\omega t)$ and $y^2 = a^2 \cos^2(\omega t)$ .

Step 3: Add the squared equations: $x^2 + y^2 = a^2 \sin^2(\omega t) + a^2 \cos^2(\omega t)$ .

Step 4: Factor out $a^2$ : $x^2 + y^2 = a^2 (\sin^2(\omega t) + \cos^2(\omega t))$ .

Step 5: Use the trigonometric identity $\sin^2(\theta) + \cos^2(\theta) = 1$ : $x^2 + y^2 = a^2$ .

Step 6: Recognize that $x^2 + y^2 = a^2$ is the equation of a circle with radius 'a' centered at the origin.

Why other options are wrong:
- Option A: An elliptical path: This would require different coefficients for the sine and cosine terms or a phase difference.
- Option C: A parabolic path: Parabolic paths are described by equations of the form $y = kx^2$ .
- Option D: A straight line path inclined equally to x and y-axis: This would require $x = y$ or $x = -y$ .

Key Concepts:
- Parametric equations
- Trigonometric identities
- Equation of a circle

Memory Aids:
- Remember the equation of a circle: $x^2 + y^2 = r^2$

Quick Tips:
- Look for trigonometric identities when dealing with parametric equations involving sine and cosine.

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Analyzing +4 | -1

A particle free to move along the

x

-axis has potential energy given as

U\left( x \right) = k\left[ {1 - \exp \left( { - {x^2}} \right)} \right]\,\,\,\,\,\,\left( {\mathrm{for} - \infty \le + \infty } \right)

Where k is a positive constant of appropriate dimensions. Then

View answer and explanation

Correct option: D

At

x = 0

, the motion of the particle is simple harmonic

Option D (At

x = 0

, the motion of the particle is simple harmonic) is correct because the potential energy function resembles that of SHM near the origin.

Solution approach

Correct Answer + Core Reason: At $x = 0$ , the motion of the particle is simple harmonic. This is because the potential energy function approximates a parabola near the origin, which is characteristic of SHM.

Step-by-Step Solution:

Step 1: Analyze the potential energy function: $U(x) = k[1 - \exp(-x^2)]$ .

Step 2: For small $x$ , we can use the Taylor series expansion of $e^{-x^2}$ around $x=0$ : $e^{-x^2} \approx 1 - x^2 + \frac{x^4}{2!} - ...$ .

Step 3: Substitute this approximation into the potential energy function: $U(x) \approx k[1 - (1 - x^2 + \frac{x^4}{2} - ...)] = k[x^2 - \frac{x^4}{2} + ...]$ .

Step 4: For small $x$ , we can neglect higher-order terms, so $U(x) \approx kx^2$ .

Step 5: This potential energy function is of the form $U(x) = \frac{1}{2} m \omega^2 x^2$ , which is the potential energy function for simple harmonic motion.

Step 6: Therefore, at $x = 0$ , the motion of the particle is simple harmonic.

Why other options are wrong:
- Option A: At points away from origin, the particle is in equilibrium. This is incorrect because the force is $F = -\frac{dU}{dx} = -k(2x e^{-x^2})$ , which is zero only at $x=0$ .
- Option B: For any finite non-zero value of $x$ , there is a force directed away from the origin. This is incorrect because the force is directed towards the origin for all non-zero $x$ .
- Option C: Its total mechanical energy is $K/2$ and it is equal to its kinetic energy at origin. This is incorrect because the potential energy at the origin is zero, so the total mechanical energy is equal to the kinetic energy at the origin, but it's not necessarily $K/2$ .

Key Concepts:
- Simple Harmonic Motion (SHM)
- Potential Energy
- Taylor Series Expansion

Memory Aids:
- Remember the Taylor series expansion for $e^x$ and $e^{-x}$ .
- Recall that SHM occurs when the potential energy is proportional to $x^2$ .

Quick Tips:
- Use Taylor series expansion to approximate functions for small values of the variable.
- Relate the potential energy function to the force acting on the particle using $F = -\frac{dU}{dx}$ .

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Understanding +4 | -1

The total energy of a simple harmonic oscillator is proportional to

View answer and explanation

Correct option: D

Square of the amplitude

Option D (Square of the amplitude) is correct because the total energy of a simple harmonic oscillator is given by

E = \frac{1}{2} k A^2

, which is proportional to the square of the amplitude.

Solution approach

Correct Answer + Core Reason: Square of the amplitude. The total energy of a simple harmonic oscillator is proportional to the square of the amplitude.

Step-by-Step Solution:

Step 1: The total energy of a simple harmonic oscillator is the sum of its kinetic and potential energies.

Step 2: The total energy can be expressed as $E = \frac{1}{2}kA^2$ , where $k$ is the spring constant and $A$ is the amplitude of the oscillation.

Step 3: From the formula, it is clear that the total energy is proportional to the square of the amplitude.

Why other options are wrong:
- Option A: Square root of displacement is incorrect because the total energy is proportional to the square of the amplitude, not the square root of displacement.
- Option B: Velocity is incorrect because the total energy is proportional to the square of the amplitude, not the velocity.
- Option C: Frequency is incorrect because the total energy is proportional to the square of the amplitude, not the frequency. The total energy is independent of frequency.

Key Concepts:
- Total energy of a simple harmonic oscillator

Memory Aids:
- Remember the formula for the total energy of a simple harmonic oscillator: $E = \frac{1}{2}kA^2$ .

Quick Tips:
- The total energy of a SHM is proportional to the square of the amplitude.

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Remembering +4 | -1

The phenomenon that occurs when the
frequency of forced vibrations on an object
matches the natural frequency of that object,
and produces a dramatic increase in amplitude
is called

View answer and explanation

Correct option: A

Resonance

Option A (Resonance) is correct because resonance occurs when the frequency of forced vibrations matches the natural frequency, leading to a large amplitude increase.

Solution approach

Correct Answer + Core Reason: Resonance. Resonance occurs when the frequency of forced vibrations matches the natural frequency.

Step-by-Step Solution:

Step 1: Define the terms:
- Forced vibration: Vibration caused by an external force.
- Natural frequency: The frequency at which an object naturally vibrates when disturbed.
- Resonance: The phenomenon where the amplitude of a vibrating object dramatically increases when the frequency of the external force matches the object's natural frequency.

Step 2: When the frequency of the forced vibrations matches the natural frequency of the object, the energy transfer is maximized, leading to a dramatic increase in amplitude. This phenomenon is called resonance.

Why other options are wrong:
- Option B: Beats is incorrect because beats occur due to the superposition of two waves with slightly different frequencies.
- Option C: Forced vibration is incorrect because it is the cause, not the phenomenon of increased amplitude.
- Option D: Damping is incorrect because damping reduces the amplitude of vibrations.

Key Concepts:
- Resonance
- Forced vibration
- Natural frequency

Memory Aids:
- Remember that resonance occurs when the driving frequency matches the natural frequency.

Quick Tips:
- Resonance leads to a large increase in amplitude.

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Application +4 | -1

A particle of mass 200 g executes SHM. The restoring force is provided by a spring of force constant 80 N/m. The time period of oscillation is

View answer and explanation

Correct option: A

0.31 s

Option A (0.31 s) is correct because it is calculated using the mass and spring constant in the formula for the period of oscillation.

Solution approach

Correct Answer + Core Reason: 0.31 s. The time period is calculated from the mass and spring constant.

Step-by-Step Solution:

Step 1: Given: mass $m = 200 \mathrm{~g} = 0.2 \mathrm{~kg}$ , spring constant $k = 80 \mathrm{~N/m}$ .

Step 2: The time period of oscillation is given by $T = 2\pi \sqrt{\frac{m}{k}}$ .

Step 3: Substitute the given values: $T = 2\pi \sqrt{\frac{0.2}{80}} = 2\pi \sqrt{\frac{1}{400}} = 2\pi \times \frac{1}{20} = \frac{\pi}{10} \approx \frac{3.14}{10} \approx 0.314 \mathrm{~s} \approx 0.31 \mathrm{~s}$ .

Why other options are wrong:
- Option B: Incorrect. This value doesn't match the correct calculation using the given mass and spring constant.
- Option C: Incorrect. This value doesn't match the correct calculation using the given mass and spring constant.
- Option D: Incorrect. This value doesn't match the correct calculation using the given mass and spring constant.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Time period of a spring-mass system: $T = 2\pi \sqrt{\frac{m}{k}}$

Memory Aids:
- Remember the formula for the time period of a spring-mass system.

Quick Tips:
- Make sure to convert all the given values to SI units before plugging them into the formulas.

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Understanding +4 | -1

If the displacement equation of a particle be represented by

y = A\sin PT + B\cos PT,

the particle executes

View answer and explanation

Correct option: C

A S.H.M.

Option C (A S.H.M.) is correct because the given equation represents the superposition of sine and cosine functions with the same frequency, which results in SHM.

Solution approach

Correct Answer + Core Reason: A S.H.M.. The equation represents a superposition of sine and cosine functions with the same frequency, resulting in SHM.

Step-by-Step Solution:

Step 1: Recognize that $y = A\sin(Pt) + B\cos(Pt)$ can be rewritten as $y = C\sin(Pt + \phi)$ , where $C = \sqrt{A^2 + B^2}$ and $\phi = \arctan(\frac{B}{A})$ .

Step 2: The equation $y = C\sin(Pt + \phi)$ represents Simple Harmonic Motion (SHM) with amplitude $C$ and angular frequency $P$ .

Why other options are wrong:
- Option A: A uniform circular motion. Incorrect, the equation represents linear motion, not circular motion.
- Option B: A uniform elliptical motion. Incorrect, the equation represents linear motion, not elliptical motion.
- Option D: A rectilinear motion. Incorrect, while it is rectilinear, it is specifically SHM.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Superposition of sine and cosine functions

Memory Aids:
- Sum of sine and cosine with same frequency is SHM.

Quick Tips:
- Recognize the general form of SHM equations.

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Application +4 | -1

A particle is executing simple harmonic motion with a period of

T

seconds and amplitude a metre. The shortest time it takes to reach a point

\frac{a}{{\sqrt 2 }}m

from its mean position in seconds is

View answer and explanation

Correct option: C

T/8

Option C (T/8) is correct because the particle's displacement is

x = a \sin(\omega t)

, and when

x = a/\sqrt{2}

,

\sin(\omega t) = 1/\sqrt{2}

, which means

\omega t = \pi/4

. Since

\omega = 2\pi/T

, we have

t = T/8

.

Solution approach

Correct Answer + Core Reason: $T/8$ . The time is calculated using the SHM equation and the given position.

Step-by-Step Solution:

Step 1: The particle starts from the mean position, so the equation of motion is $x = a \sin(\omega t)$ .

Step 2: The period is $T$ , so $\omega = \frac{2\pi}{T}$ .

Step 3: We want to find the time it takes to go from $x=0$ to $x=\frac{a}{\sqrt{2}}$ .

Step 4: Substitute $x = \frac{a}{\sqrt{2}}$ into the equation of motion: $\frac{a}{\sqrt{2}} = a \sin(\omega t)$ .

Step 5: Simplify: $\sin(\omega t) = \frac{1}{\sqrt{2}}$ .

Step 6: Solve for $\omega t$ : $\omega t = \sin^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$ .

Step 7: Solve for $t$ : $t = \frac{\pi}{4\omega} = \frac{\pi}{4(\frac{2\pi}{T})} = \frac{\pi}{4} \times \frac{T}{2\pi} = \frac{T}{8}$ .

Why other options are wrong:
- **Option A ( $T$ ): This is incorrect because it represents one full period, not the time to reach the given position.
- Option B ( $T/4$ ):** This is incorrect because it corresponds to a different position in the SHM, not $\frac{a}{\sqrt{2}}$ .
- **Option D ( $T/16$ ): This is incorrect because it's the result of a miscalculation or incorrect substitution into the SHM equation.

Key Concepts:**
- Simple Harmonic Motion
- Equation of SHM: $x = a \sin(\omega t)$
- Angular frequency: $\omega = \frac{2\pi}{T}$

Memory Aids:
- Remember the values of sine for common angles: $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ .

Quick Tips:
- Always convert the given information into the standard SHM equation.
- Pay attention to the starting position and the direction of motion.

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Application +4 | -1

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is

16\,\,cm{s^{ - 1}}

The distance of the particle from the mean position at which the speed of the particle become

8\sqrt 3 cm{s^{ - 1}},

will be

View answer and explanation

Correct option: D

2\,\,cm

Option D (2 cm) is correct because using the formula

v = \omega \sqrt{A^2 - x^2}

, we can find

\omega

and then solve for x when v =

8\sqrt{3}

cm/s.

Solution approach

Correct Answer + Core Reason: $2\,cm$ . The distance is calculated using the relationship between velocity, amplitude, and displacement in SHM.

Step-by-Step Solution:

Step 1: Given: Amplitude $A = 4 \text{ cm}$ , velocity at mean position $v_{max} = 16 \text{ cm/s}$ , and velocity at a certain distance $v = 8\sqrt{3} \text{ cm/s}$ .

Step 2: The maximum velocity in SHM is given by $v_{max} = A\omega$ , where $\omega$ is the angular frequency.

Step 3: Calculate $\omega$ : $16 = 4\omega \implies \omega = \frac{16}{4} = 4 \text{ rad/s}$ .

Step 4: The velocity at any point $x$ is given by $v = \omega \sqrt{A^2 - x^2}$ .

Step 5: Substitute the given values: $8\sqrt{3} = 4 \sqrt{4^2 - x^2}$ .

Step 6: Divide both sides by 4: $2\sqrt{3} = \sqrt{16 - x^2}$ .

Step 7: Square both sides: $(2\sqrt{3})^2 = 16 - x^2 \implies 12 = 16 - x^2$ .

Step 8: Solve for $x^2$ : $x^2 = 16 - 12 = 4$ .

Step 9: Solve for $x$ : $x = \sqrt{4} = 2 \text{ cm}$ .

Why other options are wrong:
- **Option A ( $2\sqrt{3}\,cm$ ): This is incorrect because it's the result of a miscalculation or incorrect substitution into the velocity equation.
- Option B ( $\sqrt{3}\,cm$ ): This is incorrect because it's the result of a miscalculation or incorrect substitution into the velocity equation.
- Option C ( $1\,cm$ ): This is incorrect because it's the result of a miscalculation or incorrect substitution into the velocity equation.

Key Concepts:**
- Simple Harmonic Motion
- Velocity in SHM: $v = \omega \sqrt{A^2 - x^2}$
- Maximum velocity in SHM: $v_{max} = A\omega$

Memory Aids:
- Remember the relationship between velocity, amplitude, and displacement in SHM.

Quick Tips:
- Always identify the given values and the required value.
- Use the appropriate formula for velocity in SHM.

57 Medium

57 of 94Question permalink

Understanding +4 | -1

The circular motion of a particle with constant speed is

View answer and explanation

Correct option: D

Periodic but not simple harmonic

Option D (Periodic but not simple harmonic) is correct because while the motion repeats, the restoring force is not proportional to displacement.

Solution approach

Correct Answer + Core Reason: Periodic but not simple harmonic. Circular motion repeats, but the restoring force isn't directly proportional to displacement like in SHM.

Step-by-Step Solution:

Step 1: Define periodic motion: Motion that repeats itself after a fixed interval of time.

Step 2: Define Simple Harmonic Motion (SHM): Motion where the restoring force is proportional to the displacement from the equilibrium position and acts in the opposite direction.

Step 3: Circular motion with constant speed is periodic because the particle returns to the same position after each revolution.

Step 4: However, in circular motion, the acceleration is centripetal, directed towards the center of the circle. While there is a force causing this acceleration, it is not a restoring force proportional to displacement from a fixed equilibrium point in the same way as SHM.

Why other options are wrong:
- Option A: Simple harmonic but not periodic. Incorrect, circular motion is periodic.
- Option B: Periodic and simple harmonic. Incorrect, circular motion is not SHM.
- Option C: Neither periodic nor simple harmonic. Incorrect, circular motion is periodic.

Key Concepts:
- Periodic motion
- Simple Harmonic Motion (SHM)
- Circular motion
- Centripetal acceleration

Memory Aids:
- SHM requires a restoring force proportional to displacement.

Quick Tips:
- Distinguish between periodic and SHM.

58 Medium

58 of 94Question permalink

Application +4 | -1

Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean position is separated by distance

{x_0}

({x_0} > A)

. If the maximum separation between them is

\left( {{x_0} + A} \right)

, the phase difference between their motions is

View answer and explanation

Correct option: A

\frac{\pi }{3}\;

Option A (

\frac{\pi }{3}\;

) is correct because the phase difference leads to the given maximum separation between the two particles undergoing SHM.

Solution approach

Correct Answer + Core Reason: $\frac{\pi }{3}\;$ . The phase difference is calculated based on the maximum separation condition.

Step-by-Step Solution:

Step 1: Define the positions of the two particles executing SHM.\nLet the positions of the two particles be given by:\nx1 = A \sin(ω t)\nx2 = x0 + A \sin(ω t + φ)\nwhere A is the amplitude, ω is the angular frequency, x0 is the separation of the mean positions, and φ is the phase difference.

Step 2: Find the separation between the particles.\nThe separation between the particles is:\nx = x2 - x1 = x0 + A \sin(ω t + φ) - A \sin(ω t)

Step 3: Use trigonometric identity to simplify the expression.\nUsing the trigonometric identity \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b), we have:\nx = x0 + A [\sin(ω t)\cos(φ) + \cos(ω t)\sin(φ)] - A \sin(ω t)\nx = x0 + A \sin(ω t) [\cos(φ) - 1] + A \cos(ω t) \sin(φ)

Step 4: Determine the condition for maximum separation.\nThe maximum separation is given as x0 + A. To find the maximum value of x, we need to find the maximum value of the time-dependent part.\nx - x0 = A \sin(ω t) [\cos(φ) - 1] + A \cos(ω t) \sin(φ)\nFor the maximum separation to be x0 + A, the maximum value of (x - x0) must be A.\nThus, the maximum value of A \sin(ω t) [\cos(φ) - 1] + A \cos(ω t) \sin(φ) must be A.\nThis can be written as:\nA $\sqrt{[\cos(φ) - 1]^2 + [\sin(φ)]^2}$ = A
$\sqrt{[\cos^2(φ) - 2\cos(φ) + 1] + \sin^2(φ)}$ = 1
$\sqrt{1 - 2\cos(φ) + 1}$ = 1
$\sqrt{2 - 2\cos(φ)}$ = 1
2 - 2\cos(φ) = 1
2\cos(φ) = 1
\cos(φ) = $\frac{1}{2}$
φ = $\frac{π}{3}$

Why other options are wrong:
- Option B: $\frac{\pi }{4}$ This phase difference would not result in the given maximum separation.
- Option C: $\frac{\pi }{6}$ This phase difference would not result in the given maximum separation.
- Option D: $\frac{\pi }{2}$ This phase difference would not result in the given maximum separation.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase difference
- Trigonometric identities

Memory Aids:
- Remember the trigonometric identity \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)

Quick Tips:
- Always define the positions of the particles first.
- Use trigonometric identities to simplify the expression.
- Determine the condition for maximum or minimum separation.

59 Medium

59 of 94Question permalink

Understanding +4 | -1

The path difference between the two waves:

{y_1} = {a_1}\,\sin \left( {\omega t - \frac{{2\pi x}}{\lambda }} \right)\,

and

{y_2} = {a_2}\,\sin \left( {\omega t - \frac{{2\pi x}}{\lambda } + \phi } \right)\,

will be

View answer and explanation

Correct option: C

\frac{\lambda }{{2\pi }}\phi

Option C (

\frac{\lambda }{{2\pi }}\phi

) is correct because the path difference is related to the phase difference by the formula: path difference = (wavelength / 2π) * phase difference.

Solution approach

Correct Answer + Core Reason: $\frac{\lambda }{{2\pi }}\phi$ . The path difference is calculated using the relationship between phase difference and path difference.

Step-by-Step Solution:

Step 1: Identify the given information: The two waves are given by ${y_1} = {a_1}\,\sin \left( {\omega t - \frac{{2\pi x}}{\lambda }} \right)\, and$ {y_2} = {a_2}\,\sin \left( {ω t - $\frac{{2π x}}{λ }$ + φ } \right)\, $. The phase difference between the two waves is$ φ $.\n\n**Step 2:** Recall the relationship between path difference ($ Δ x $) and phase difference ($ Δ φ $):$ Δ φ = $\frac{{2π }}{λ }$ Δ x $\n\n**Step 3:** Rearrange the formula to solve for the path difference:$ Δ x = $\frac{λ }{{2π }}$ Δ φ $\n\n**Step 4:** Substitute the given phase difference$ φ $into the equation:$ Δ x = $\frac{λ }{{2π }}$ φ $\n\n**Step 5:** The path difference is$ $\frac{λ }{{2π }}$ φ $.\n\n**Why other options are wrong:**\n- **Option A:**$ $\frac{{2π }}{λ }$ φ $is incorrect because it represents the phase difference in terms of path difference, not the path difference itself.\n- **Option B:**$ $\frac{{2π }}{λ }$ \left( {φ - $\frac{π }{2}$ } \right) $is incorrect because it includes an additional phase shift of$ $\frac{π }{2}$ $which is not present in the given problem.\n- **Option D:**$ $\frac{{2π }}{λ }$ \left( {φ + $\frac{π }{2}$ } \right) $is incorrect because it includes an additional phase shift of$ $\frac{π }{2}$ $which is not present in the given problem.\n\n**Key Concepts:**\n- Wave motion\n- Phase difference\n- Path difference\n\n**Memory Aids:**\n- Remember the relationship: Phase difference = (2π / wavelength) * Path difference\n\n**Quick Tips:**\n- Always identify the phase difference correctly from the given wave equations.\n- Use the correct formula relating phase difference and path difference.$

60 Medium

60 of 94Question permalink

Analyzing +4 | -1

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of

{10^{ - 2}}m

. The relative change in the angular frequency of the pendulum is best given by

View answer and explanation

Correct option: C

{10^{ - 3}}rad\mathrm{/}s

Option C (

{10^{ - 3}}rad\mathrm{/}s

) is correct because the relative change in angular frequency is related to the amplitude and angular frequency of the support's oscillation.

Solution approach

Correct Answer + Core Reason: ${10^{ - 3}}rad\mathrm{/}s$ . The relative change in angular frequency is calculated based on the given parameters.

Step-by-Step Solution:

Step 1: Identify the given parameters: length of the pendulum $L = 1 \text{ m}$ , angular frequency of the pendulum $\omega = 10 \text{ rad/s}$ , angular frequency of the support $\omega_s = 1 \text{ rad/s}$ , amplitude of the support $A = 10^{-2} \text{ m}$ .

Step 2: The effective acceleration due to the oscillating support is $a_{eff} = A \omega_s^2 = (10^{-2} \text{ m})(1 \text{ rad/s})^2 = 10^{-2} \text{ m/s}^2$ .

Step 3: The change in the effective gravitational acceleration is $\Delta g = a_{eff} = 10^{-2} \text{ m/s}^2$ .

Step 4: The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g}{L}}$ .

Step 5: The change in angular frequency is given by $\Delta \omega = \frac{1}{2} \sqrt{\frac{L}{g}} \frac{\Delta g}{L} = \frac{1}{2} \frac{\Delta g}{\sqrt{gL}} = \frac{1}{2} \frac{10^{-2}}{\sqrt{10 \times 1}} = \frac{10^{-2}}{2\sqrt{10}} \approx \frac{10^{-2}}{2 \times 3.16} \approx 1.58 \times 10^{-3}$ .

Step 6: The relative change in angular frequency is $\frac{\Delta \omega}{\omega} = \frac{1.58 \times 10^{-3}}{10} \approx 1.58 \times 10^{-4}$ . Therefore, $\Delta \omega = \frac{1}{2} \frac{\Delta g}{\sqrt{gL}} = \frac{1}{2} \frac{10^{-2}}{\sqrt{10}} \approx 0.00158 \text{ rad/s}$ .\nSince the question asks for the relative change, we calculate $\Delta \omega \approx 10^{-3} \text{ rad/s}$ .

Why other options are wrong:
- Option A: $1\,rad\mathrm{/}s$ : This is incorrect as it is much larger than the calculated change.
- Option B: ${10^{ - 5}}\,rad\mathrm{/}s$ : This is incorrect as it is much smaller than the calculated change.
- Option D: ${10^{ - 1}}\,rad\mathrm{/}s$ : This is incorrect as it is larger than the calculated change.

Key Concepts:
- Simple pendulum
- Angular frequency
- Effective acceleration

Memory Aids:
- $\omega = \sqrt{\frac{g}{L}}$

Quick Tips:
- Remember to consider the effective acceleration when the support is oscillating.

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