OSCILLATIONS MCQs for NEET UG PHYSICS — Mixed difficulty

1 Medium

Application +4 | -1

A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is

{f_1}

. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is

{f_2}

. Which of the following relations between the frequencies is correct

View answer and explanation

Correct option: D

{f_2} = \sqrt 2 {f_1}

Option D (

{f_2} = \sqrt 2 {f_1}

) is correct because cutting the spring in half doubles the spring constant, which increases the frequency by a factor of

\sqrt{2}

.

Solution approach

Correct Answer + Core Reason: ${f_2} = \sqrt 2 {f_1}$ . Cutting the spring in half doubles the spring constant.

Step-by-Step Solution:

Step 1: The frequency of vibration of a mass-spring system is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$ , where $k$ is the spring constant and $m$ is the mass.

Step 2: When the spring is cut into two equal parts, the spring constant of each part becomes $2k$ . This is because $k \propto \frac{1}{l}$ , where $l$ is the length of the spring. So, if the length is halved, the spring constant is doubled.

Step 3: The new frequency $f_2$ is given by $f_2 = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$ .

Step 4: Divide the new frequency by the original frequency: $\frac{f_2}{f_1} = \frac{\frac{1}{2\pi} \sqrt{\frac{2k}{m}}}{\frac{1}{2\pi} \sqrt{\frac{k}{m}}} = \sqrt{\frac{2k}{m} \times \frac{m}{k}} = \sqrt{2}$ .

Step 5: Therefore, $f_2 = \sqrt{2} f_1$ .

Why other options are wrong:
- Option A: Incorrect. The new frequency is $\sqrt{2}$ times the original frequency, not $\frac{1}{\sqrt{2}}$ times.
- Option B: Incorrect. The frequency changes because the spring constant changes.
- Option C: Incorrect. The new frequency is $\sqrt{2}$ times the original frequency, not 2 times.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Frequency of oscillation of a spring-mass system: $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$
- Spring constant and length relationship: $k \propto \frac{1}{l}$

Memory Aids:
- Remember that cutting a spring in half doubles its spring constant.

Quick Tips:
- When a spring is cut into $n$ equal parts, the spring constant of each part becomes $nk$ .

2 Medium

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Understanding +4 | -1

The time period of a loaded spring on earth is 6s. On the surface of moon, the time period of the same loaded spring will be.

View answer and explanation

Correct option: D

6 s

Option D (6 s) is correct because the time period of a spring-mass system depends on the mass and spring constant, which remain the same on the Moon.

Solution approach

Correct Answer + Core Reason: 6 s. The time period of a spring-mass system is independent of gravity.

Step-by-Step Solution:

Step 1: Recall the formula for the time period of a spring-mass system: $T = 2\pi \sqrt{\frac{m}{k}}$ , where $m$ is the mass and $k$ is the spring constant.

Step 2: The time period depends only on the mass and the spring constant, which are the same on Earth and the Moon.

Step 3: Therefore, the time period remains the same, which is 6 s.

Why other options are wrong:
- Option A: $\sqrt 6 \,s$ : This is incorrect because the time period is independent of gravity.
- Option B: $6\sqrt 6 \,s$ : This is incorrect because the time period is independent of gravity.
- Option C: 1 s: This is incorrect because the time period is independent of gravity.

Key Concepts:
- Spring-mass system
- Time period of a spring-mass system

Memory Aids:
- The time period of a spring-mass system depends only on the mass and spring constant.

Quick Tips:
- Remember that the time period is independent of gravity.

3 Easy

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Analyzing +4 | -1

Under the action of a force

F=-k x^3

, the motion of a particle is

(k=

a positive constant

)

View answer and explanation

Correct option: D

periodic but not simple harmonic

Option D (periodic but not simple harmonic) is correct because the force is proportional to

x^3

, not

x

, which is required for SHM, but the motion will still repeat.

Solution approach

Correct Answer + Core Reason: periodic but not simple harmonic. The force is not proportional to $x$ , but the motion is still periodic.

Step-by-Step Solution:

Step 1: Analyze the given force: $F = -kx^3$ .

Step 2: Recall the condition for Simple Harmonic Motion (SHM): The restoring force must be proportional to the displacement from the equilibrium position, i.e., $F = -kx$ , where $k$ is a constant.

Step 3: Since the given force is proportional to $x^3$ and not $x$ , the motion is not SHM.

Step 4: However, the force is still a restoring force (it acts in the opposite direction to the displacement), so the motion will be periodic. The particle will oscillate around the equilibrium position.

Why other options are wrong:
- Option A (simple harmonic motion): This is incorrect because the force is not proportional to $x$ .
- Option B (uniformly accelerated motion): This is incorrect because the acceleration is not constant; it depends on $x^3$ .
- Option C (not periodic): This is incorrect because the restoring force will cause the particle to oscillate, making the motion periodic.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Periodic motion
- Restoring force

Memory Aids:
- Remember that for SHM, the restoring force must be proportional to the displacement from the equilibrium position.

Quick Tips:
- Always check the relationship between the force and displacement to determine if the motion is SHM.

4 Medium

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Analyzing +4 | -1

S-I: A particle executes SHM with frequency

f

. The frequency with which total mechanical energy of particle oscillates is

2 f

.
S-II: Both kinetic as well as potential energy of particle in SHM oscillate with a frequency

2 f.

View answer and explanation

Correct option: D

S-I is false

\&

S-II is true

Option D (S-I is false & S-II is true) is correct because the frequency of oscillation of total mechanical energy is constant, while kinetic and potential energies oscillate at twice the frequency of SHM.

Solution approach

Correct Answer + Core Reason: S-I is false $\& S-II is true. The total mechanical energy is constant, while kinetic and potential energies oscillate at twice the frequency of SHM.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Statement I: A particle executes SHM with frequency$ f $. The frequency with which total mechanical energy of particle oscillates is$ 2f $. This is false. The total mechanical energy in SHM is constant, so it doesn't oscillate. Therefore, its frequency of oscillation is 0.\n\n**Step 2:** Statement II: Both kinetic as well as potential energy of particle in SHM oscillate with a frequency$ 2f $. This is true. The kinetic and potential energies vary sinusoidally with time, and their frequency is twice the frequency of the SHM.\n\n**Why other options are wrong:**\n- **Option A:** S-I$ \& S-II are true. This is incorrect because Statement I is false.
- Option B: S-I $\& S-II are false. This is incorrect because Statement II is true.\n- **Option C:** S-I is true$ \& S-II is false. This is incorrect because Statement I is false and Statement II is true.

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

Memory Aids:
- Remember that the total mechanical energy in SHM is constant.
- Remember that the kinetic and potential energies oscillate with twice the frequency of SHM.

Quick Tips:
- Understand the energy variations in SHM.
- Know that total energy is constant, while kinetic and potential energies oscillate.

5 Hard

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Application +4 | -1

The phase difference (in radian) between two SHM whose equations are

y_1=10 \sin \left(10 \pi t+\frac{\pi}{3}\right)

and

y_2=12 \sin \left(8 \pi t+\frac{\pi}{4}\right)

at

t=0.5 \mathrm{~s}

is:

View answer and explanation

Correct option: B

\frac{13 \pi}{12}

Option B (

\frac{{13 \pi}}{{12}}

) is correct because it represents the phase difference calculated at t=0.5s between the two given SHM equations.

Solution approach

Correct Answer + Core Reason: $\frac{{13 \pi}}{{12}}$ . The phase difference is calculated by substituting t=0.5s into the arguments of the sine functions and finding the difference.

Step-by-Step Solution:

Step 1: Identify the given equations: $y_1 = 10 \sin(10 \pi t + \frac{\pi}{3})$ and $y_2 = 12 \sin(8 \pi t + \frac{\pi}{4})$ .

Step 2: Calculate the phase of $y_1$ at $t = 0.5 \text{{s}}$ : $\phi_1 = 10 \pi (0.5) + \frac{\pi}{3} = 5\pi + \frac{\pi}{3} = \frac{{16\pi}}{3}$ .

Step 3: Calculate the phase of $y_2$ at $t = 0.5 \text{{s}}$ : $\phi_2 = 8 \pi (0.5) + \frac{\pi}{4} = 4\pi + \frac{\pi}{4} = \frac{{17\pi}}{4}$ .

Step 4: Find the phase difference: $\Delta \phi = |\phi_1 - \phi_2| = |\frac{{16\pi}}{3} - \frac{{17\pi}}{4}| = |\frac{{64\pi - 51\pi}}{12}| = |\frac{{13\pi}}{12}| = \frac{{13\pi}}{12}$ .

Why other options are wrong:
- **Option A ( $\frac{{11 \pi}}{{12}}$ ): This is incorrect because it's not the correct phase difference calculated at t=0.5s.
- Option C ( $\pi$ ): This is incorrect because it's not the correct phase difference calculated at t=0.5s.
- Option D ( $\frac{{17 \pi}}{{12}}$ ): This is incorrect because it's not the correct phase difference calculated at t=0.5s. The absolute value of the difference must be taken.

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

Memory Aids:
- Phase is the argument of the sine or cosine function.

Quick Tips:**
- Always substitute the given time into the phase equations before finding the difference.
- Remember to take the absolute value of the phase difference.

6 Easy

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Analyzing +4 | -1

A vertical spring block system is made to oscillate.
Statement-1: Its time period on earth is more than that on the moon.
Statement-2: Its extension on earth (in equilibrium) is more than that on the moon.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Statement-1 is false, Statement-2 is true. The time period of a spring-block system is independent of gravity, while the extension in equilibrium depends on gravity.

Solution approach

Correct Answer + Core Reason: Statement-1 is false, Statement-2 is true. Time period is independent of gravity, but extension depends on it.

Step-by-Step Solution:

Step 1: Write the expression for the time period of a vertical spring-block system: $T = 2 \pi \sqrt{\frac{m}{k}}$ , where $m$ is mass and $k$ is the spring constant.

Step 2: Observe that the time period is independent of the acceleration due to gravity ( $g$ ). Therefore, the time period on Earth is the same as the time period on the Moon.

Step 3: Write the expression for the extension in equilibrium: $x = \frac{{mg}}{k}$ .

Step 4: Since the acceleration due to gravity on the Moon is less than that on Earth, the extension on Earth is more than that on the Moon.

Step 5: Conclude that Statement-1 is false and Statement-2 is true.

Why other options are wrong:
- Option A: Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1. This is incorrect because Statement-1 is false.
- Option B: Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1. This is incorrect because Statement-1 is false.
- Option C: Statement-1 is true, Statement-2 is false. This is incorrect because both statements are not true.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Time Period of a Spring-Mass System
- Extension in Equilibrium

Memory Aids:
- $T = 2 \pi \sqrt{\frac{m}{k}}$ (Time period of a spring-mass system)
- $x = \frac{{mg}}{k}$ (Extension in equilibrium)

Quick Tips:
- The time period of a spring-mass system is independent of gravity.
- The extension in equilibrium depends on gravity.

7 Medium

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Application +4 | -1

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is

U\left( x \right) = k{\left[ x \right]^3}

, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

View answer and explanation

Correct option: A

Proportional to

\frac{1}{{\sqrt a }}

Option A (Proportional to

\frac{1}{{\sqrt a }}

) is correct because the time period of oscillation is inversely proportional to the square root of the amplitude when the potential energy is proportional to

x^3

.

Solution approach

Correct Answer + Core Reason: Proportional to $\frac{1}{{\sqrt a }}$ . The time period is inversely proportional to the square root of the amplitude.

Step-by-Step Solution:

Step 1: Given potential energy $U(x) = k[x]

8 Medium

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Analyzing +4 | -1

A particle moves according to the law,

x = rcos\frac{{\pi t}}{{2.}}

The distance covered by it the time interval between t= 0 to t=3s is

View answer and explanation

Correct option: C

3r

Option C (3r) is correct because the particle travels a distance of 'r' from t=0 to t=1, then returns 'r' from t=1 to t=2, and then travels another 'r' from t=2 to t=3, resulting in a total distance of 3r.

Solution approach

Correct Answer + Core Reason: $3r$ . The distance covered is the sum of the absolute values of the displacements between the given times.

Step-by-Step Solution:

Step 1: Determine the position at $t=0$ : $x(0) = r \cos(\frac{\pi (0)}{2}) = r \cos(0) = r$ .

Step 2: Determine the position at $t=1$ : $x(1) = r \cos(\frac{\pi (1)}{2}) = r \cos(\frac{\pi}{2}) = 0$ .

Step 3: Determine the position at $t=2$ : $x(2) = r \cos(\frac{\pi (2)}{2}) = r \cos(\pi) = -r$ .

Step 4: Determine the position at $t=3$ : $x(3) = r \cos(\frac{\pi (3)}{2}) = r \cos(\frac{{3\pi}}{2}) = 0$ .

Step 5: Calculate the distance from $t=0$ to $t=1$ : $|x(1) - x(0)| = |0 - r| = r$ .

Step 6: Calculate the distance from $t=1$ to $t=2$ : $|x(2) - x(1)| = |-r - 0| = r$ .

Step 7: Calculate the distance from $t=2$ to $t=3$ : $|x(3) - x(2)| = |0 - (-r)| = r$ .

Step 8: Calculate the total distance: $r + r + r = 3r$ .

Why other options are wrong:
- **Option A ( $r$ ): This is incorrect because it only considers the distance from t=0 to t=1, but not the subsequent movements.
- Option B ( $2r$ ): This is incorrect because it doesn't account for the movement from t=2 to t=3.
- Option D ( $4r$ ): This is incorrect as the particle doesn't cover a distance of 4r in the given time interval.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Distance vs. Displacement

Memory Aids:
- Distance is the total path length, while displacement is the change in position.

Quick Tips:**
- Pay attention to the difference between distance and displacement. In SHM, the distance can be greater than the amplitude.

9 Medium

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Understanding +4 | -1

(A): Since and cosine functions are periodic functions
(R): Sinusoidal functions repeat its values after a definite interval of time

View answer and explanation

Correct option: A

If Both A

\&

R are true and R is correct explanation of A

Option A (If Both A & R are true and R is correct explanation of A) is correct because both the assertion and the reason are true, and the reason correctly explains why sine and cosine functions are periodic.

Solution approach

Correct Answer + Core Reason: If Both A $\& R are true and R is correct explanation of A. Both the assertion and reason are true, and the reason correctly explains the assertion.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Analyze the Assertion (A): Sine and cosine functions are periodic functions. This statement is true because both sine and cosine functions repeat their values after a fixed interval.\n\n**Step 2:** Analyze the Reason (R): Sinusoidal functions repeat their values after a definite interval of time. This statement is also true and explains why sine and cosine functions are periodic.\n\n**Step 3:** Determine if R is the correct explanation for A: The reason (R) directly explains why the assertion (A) is true. The periodicity of sine and cosine functions is due to their repeating values after a definite interval of time.\n\n**Why other options are wrong:**\n- **Option B:** If Both A$ \& R are true but R is not correct explanation of A: This is incorrect because R is the correct explanation for A.
- Option C: If A is true but R is false: This is incorrect because both A and R are true.
- Option D: If both A $\& R are false: This is incorrect because both A and R are true.\n\n**Key Concepts:**\n- Periodic functions\n- Sine and cosine functions\n- Periodicity\n\n**Memory Aids:**\n- Remember that periodic functions repeat their values after a fixed interval.\n\n**Quick Tips:**\n- Understand the definitions of assertion and reason questions.\n- Check if both statements are true and if the reason correctly explains the assertion.$

10 Medium

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Application +4 | -1

Velocity of a simple harmonic oscillation is given by

V = \sqrt {144 - 16{x^2}} {\mkern 1mu} {\mkern 1mu} m{s^{ - 1}}

where x is displacement in metre. Then maximum velocity of that oscillator is

View answer and explanation

Correct option: B

12\,\,m{s^{ - 1}}

Option B (

12\,m{s^{ - 1}}

) is correct because the maximum velocity of the oscillator is found by comparing the given equation with the standard form of velocity in SHM.

Solution approach

Correct Answer + Core Reason: $12\,m{s^{ - 1}}$ . The maximum velocity is found by comparing the given equation with the standard form.

Step-by-Step Solution:

Step 1: Identify the given equation for velocity: $V = \sqrt{144 - 16x^2} \, m/s$ .

Step 2: Rewrite the equation in a more recognizable form. Factor out 16 from under the square root: $V = \sqrt{16(9 - x^2)} = 4\sqrt{9 - x^2} = 4\sqrt{3^2 - x^2}$ .

Step 3: Recall the standard form of velocity in SHM: $v = \omega \sqrt{A^2 - x^2}$ , where $\omega$ is the angular frequency and $A$ is the amplitude.

Step 4: Compare the given equation with the standard form. We have $\omega = 4$ and $A = 3$ . Therefore, $V = 4\sqrt{3^2 - x^2}$ .

Step 5: The maximum velocity occurs when $x = 0$ . So, $V_{max} = \omega A = 4 \times 3 = 12 \, m/s$ .

Why other options are wrong:
- Option A: $6\,m{s^{ - 1}}$ is incorrect as it does not match the correct calculation.
- Option C: $16\,m{s^{ - 1}}$ is incorrect as it does not match the correct calculation.
- Option D: $8\,m{s^{ - 1}}$ is incorrect as it does not match the correct calculation.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Velocity in SHM
- Amplitude
- Angular frequency

Memory Aids:
- Remember the formula for velocity in SHM: $v = \omega \sqrt{A^2 - x^2}$
- Maximum velocity occurs at the equilibrium position (x=0) and is given by $v_{max} = \omega A$

Quick Tips:
- Recognize the standard form of equations in SHM to easily identify parameters like amplitude and angular frequency.
- The maximum velocity occurs when the displacement is zero.

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Remembering +4 | -1

Which one of the following is the correct relation between frequency

\mathrm{f}

and angular frequency

\omega

?

View answer and explanation

Correct option: B

\omega=2 \pi \mathrm{f}

Option B (

\omega=2 \pi \mathrm{f}

) is correct because it is the fundamental relationship between angular frequency and frequency.

Solution approach

Correct Answer + Core Reason: $\omega=2 \pi \mathrm{f}$ . This is the correct relationship between angular frequency and frequency.

Step-by-Step Solution:

Step 1: Recall the definition of frequency (f): It is the number of oscillations per unit time (usually seconds).

Step 2: Recall the definition of angular frequency ( $\omega$ ): It is the rate of change of the angle of an object moving on a circular path, or the rate of change of the phase of a sinusoidal waveform.

Step 3: Relate the two quantities. One complete oscillation corresponds to an angle of $2\pi$ radians. Therefore, the angular frequency is $2\pi$ times the frequency: $\omega = 2\pi f$ .

Why other options are wrong:
- **Option A ( $\mathrm{f}=\pi \omega$ ):** This is incorrect because it has the wrong relationship between f and $\omega$ .
- **Option C ( $\mathrm{f}=2 \omega / \pi$ ):** This is incorrect because it has the wrong relationship between f and $\omega$ .
- **Option D ( $\mathrm{f}=2 \pi \omega$ ):** This is incorrect because it has the wrong relationship between f and $\omega$ .

Key Concepts:
- Frequency
- Angular Frequency
- Relationship between Frequency and Angular Frequency

Memory Aids:
- Remember the formula $\omega = 2\pi f$ .

Quick Tips:
- Always use the correct relationship between frequency and angular frequency when solving problems involving oscillations.

12 Medium

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Analyzing +4 | -1

The bob of a pendulum is in the lowest position and has a horizontal velocity of

\sqrt {gL}

where

L

is length of the pendulum and

g

is acceleration due to gravity is given to it. Then the bob

View answer and explanation

Correct option: C

Oscillates about the lowest position as mean position

Option C (Oscillates about the lowest position as mean position) is correct because the initial velocity is not high enough for the bob to complete a full circle.

Solution approach

Correct Answer + Core Reason: Oscillates about the lowest position as mean position. The bob oscillates because the initial velocity is less than the velocity required to complete a vertical circle.

Step-by-Step Solution:

Step 1: The minimum velocity required to complete a vertical circle at the lowest point is $v = \sqrt{5gL}$ .

Step 2: The given velocity is $v = \sqrt{gL}$ .

Step 3: Since $\sqrt{gL} < \sqrt{5gL}$ , the bob will not complete a vertical circle.

Step 4: The bob will rise to a certain height and then oscillate about the lowest position as the mean position.

Why other options are wrong:
- Option A: Just completes a vertical circle. This is incorrect because the initial velocity is less than the required velocity.
- Option B: Just completes semi-circle. This is incorrect because the initial velocity is not sufficient to reach the top of the circle (semi-circle).
- Option D: Oscillates as conceal pendulum. This is incorrect because the pendulum oscillates in a vertical plane, not as a conical pendulum.

Key Concepts:
- Vertical Circular Motion
- Simple Pendulum
- Conservation of Energy

Memory Aids:
- Remember the minimum velocity required to complete a vertical circle at the lowest point: $v = \sqrt{5gL}$ .

Quick Tips:
- Compare the given velocity with the minimum velocity required to complete a vertical circle.
- If the given velocity is less than the minimum velocity, the bob will oscillate.

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Remembering +4 | -1

The differential equation of a particle executing simple harmonic motion along

y

-axis is

View answer and explanation

Correct option: A

\frac{d^2 y}{d t^2}+\omega^2 y=0

Option A (

\frac{{d^2 y}}{{d t^2}}+\omega^2 y=0

) is correct because it represents the standard differential equation for SHM.

Solution approach

Correct Answer + Core Reason: $\frac{{d^2 y}}{{d t^2}}+\omega^2 y=0$ . This is the standard differential equation for SHM.

Step-by-Step Solution:

Step 1: Recall the general equation for SHM: $y(t) = A \cos(\omega t + \phi)$ , where A is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant.

Step 2: Find the first derivative of y with respect to time: $\frac{{dy}}{{dt}} = -A\omega \sin(\omega t + \phi)$ .

Step 3: Find the second derivative of y with respect to time: $\frac{{d^2y}}{{dt^2}} = -A\omega^2 \cos(\omega t + \phi)$ .

Step 4: Substitute the original equation $y(t) = A \cos(\omega t + \phi)$ into the second derivative equation: $\frac{{d^2y}}{{dt^2}} = -\omega^2 y$ .

Step 5: Rearrange the equation to get the standard form of the differential equation for SHM: $\frac{{d^2y}}{{dt^2}} + \omega^2 y = 0$ .

Why other options are wrong:
- **Option B ( $\frac{{d^2 y}}{{d t^2}}+\omega^2 y^2=0$ ):** This is incorrect because it has a $y^2$ term, which is not part of the standard SHM differential equation.
- **Option C ( $\frac{{d^2 y}}{{d t^2}}-\omega^2 y=0$ ):** This is incorrect because it has a negative sign in front of the $\omega^2 y$ term. This would represent exponential growth or decay, not oscillation.
- **Option D ( $\frac{{d y}}{{d t}}+\omega y=0$ ): This is incorrect because it's a first-order differential equation, not a second-order one, and it doesn't represent SHM.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Differential Equation of SHM

Memory Aids:**
- Remember the standard form of the differential equation for SHM: $\frac{{d^2y}}{{dt^2}} + \omega^2 y = 0$ .

Quick Tips:
- The differential equation for SHM is a second-order linear homogeneous differential equation with constant coefficients.

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Analyzing +4 | -1

The displacement

y

of a particle is given by

y=4 \cos ^2\left(\frac{t}{2}\right) \sin (1000 t)

. This expression may be considered to be a result of the superposition of how many simple harmonic motion

View answer and explanation

Correct option: A

3

Option A (3) is correct because the given expression can be simplified using trigonometric identities to a superposition of three simple harmonic motions.

Solution approach

Correct Answer + Core Reason: 3. The given expression can be simplified to a sum of three SHM terms using trigonometric identities.

Step-by-Step Solution:

Step 1: Use the trigonometric identity $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ to rewrite the given expression:
$y = 4 \cos^2(\frac{t}{2}) \sin(1000t) = 4 \left( \frac{1 + \cos(t)}{2} \right) \sin(1000t) = 2(1 + \cos(t)) \sin(1000t)$ .

Step 2: Expand the expression: $y = 2\sin(1000t) + 2\cos(t)\sin(1000t)$ .

Step 3: Use the trigonometric identity $2\cos(A)\sin(B) = \sin(A+B) - \sin(A-B)$ to rewrite the second term:
$2\cos(t)\sin(1000t) = \sin(1000t + t) + \sin(1000t - t) = \sin(1001t) - \sin(-999t) = \sin(1001t) + \sin(999t)$ .

Step 4: Substitute this back into the expression for $y$ : $y = 2\sin(1000t) + \sin(1001t) + \sin(999t)$ .

Step 5: Count the number of simple harmonic motions: There are three terms, each representing a simple harmonic motion with different frequencies (1000, 1001, and 999).

Why other options are wrong:
- Option B (0): This is incorrect because the expression can be simplified into three SHM terms.
- Option C (0): This is incorrect because the expression can be simplified into three SHM terms.
- Option D (0): This is incorrect because the expression can be simplified into three SHM terms.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Trigonometric Identities
- Superposition of SHM

Memory Aids:
- Remember the trigonometric identities for $\cos^2(x)$ and $2\cos(A)\sin(B)$ .

Quick Tips:
- Use trigonometric identities to simplify complex expressions into a sum of simple harmonic motions.

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Application +4 | -1

An object on a spring vibrates in a simple harmonic motion at a frequency of 4 hertz and an amplitude of 8 cm. If the mass of the object is 200 g, the spring constant is

View answer and explanation

Correct option: C

126 N/m

Option C (126 N/m) is correct because it is calculated using the frequency and mass to find the spring constant.

Solution approach

Correct Answer + Core Reason: 126 N/m. The spring constant is calculated from the frequency and mass.

Step-by-Step Solution:

Step 1: Given: frequency $f = 4 \mathrm{~Hz}$ , amplitude $A = 8 \mathrm{~cm} = 0.08 \mathrm{~m}$ , mass $m = 200 \mathrm{~g} = 0.2 \mathrm{~kg}$ .

Step 2: The angular frequency is $\omega = 2\pi f = 2\pi (4) = 8\pi \mathrm{~rad/s}$ .

Step 3: The spring constant $k$ is related to the angular frequency and mass by $\omega = \sqrt{\frac{k}{m}}$ .

Step 4: Solve for $k$ : $k = m\omega^2 = 0.2 \times (8\pi)^2 = 0.2 \times 64 \pi^2 \approx 0.2 \times 64 \times 9.87 \approx 126.336 \mathrm{~N/m} \approx 126 \mathrm{~N/m}$

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

Key Concepts:
- Simple Harmonic Motion (SHM)
- Angular frequency: $\omega = 2\pi f$
- Relationship between spring constant, mass, and angular frequency: $\omega = \sqrt{\frac{k}{m}}$

Memory Aids:
- Remember the formula $\omega = \sqrt{\frac{k}{m}}$ .

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

16 Medium

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Understanding +4 | -1

The average acceleration of a particle performing SHM over one complete oscillation is

View answer and explanation

Correct option: C

Zero

Option C (Zero) is correct because the velocity changes direction over the oscillation, resulting in zero average acceleration.

Solution approach

Correct Answer + Core Reason: Zero. Over a complete oscillation, the velocity changes direction, resulting in zero average acceleration.

Step-by-Step Solution:

Step 1: Define average acceleration: Average acceleration is the change in velocity divided by the change in time: $a_{avg} = \frac{\Delta v}{\Delta t}$ .

Step 2: In SHM, after one complete oscillation, the particle returns to its starting point with the same speed but in the opposite direction (if starting at an extreme) or the same speed and direction (if starting at the mean position).

Step 3: If the particle starts at an extreme position, its velocity at the beginning is 0 and its velocity after one oscillation is also 0. If the particle starts at the mean position, its velocity is +v and after one oscillation, its velocity is also +v. In either case, the *change* in velocity over one full oscillation, considering direction, is zero.

Step 4: Therefore, the average acceleration over one complete oscillation is zero.

Why other options are wrong:
- Option A: $\frac{{\omega ^2}A}{2}$ . Incorrect, this is not the average acceleration.
- Option B: $\frac{{\omega ^2}A}{{\sqrt 2 }}$ . Incorrect, this is not the average acceleration.
- Option D: $A{\omega ^2}$ . Incorrect, this is the maximum acceleration, not the average.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Average acceleration
- Oscillation

Memory Aids:
- Average acceleration over a cycle is zero.

Quick Tips:
- Consider the direction of velocity when calculating average acceleration.

17 Medium

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Application +4 | -1

The time period of a particle performing linear SHM is 18s. What is the time taken by it to make a displacement equal to half its amplitude from its equilibrium position ?

View answer and explanation

Correct option: C

\frac{3}{2}s

Option C (

\frac{3}{2}s

) is correct because it is calculated using the SHM equation and the given time period.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}s$ . The time taken to reach half the amplitude is calculated using the SHM equation and the given time period.

Step-by-Step Solution:

Step 1: Define the SHM equation: $x = A \sin(\omega t)$ , where $x$ is the displacement, $A$ is the amplitude, $\omega$ is the angular frequency, and $t$ is the time.

Step 2: We are given that $x = \frac{A}{2}$ . Substitute this into the equation: $\frac{A}{2} = A \sin(\omega t)$ .

Step 3: Simplify the equation: $\frac{1}{2} = \sin(\omega t)$ .

Step 4: Find the angle whose sine is $\frac{1}{2}$ : $\omega t = \frac{\pi}{6}$ .

Step 5: We know that $\omega = \frac{2\pi}{T}$ , where $T$ is the time period. Given $T = 18s$ , we have $\omega = \frac{2\pi}{18} = \frac{\pi}{9}$ .

Step 6: Substitute the value of $\omega$ into the equation $\omega t = \frac{\pi}{6}$ : $\frac{\pi}{9} t = \frac{\pi}{6}$ .

Step 7: Solve for $t$ : $t = \frac{\pi}{6} \times \frac{9}{\pi} = \frac{9}{6} = \frac{3}{2}s$ .

Final Answer: $t = \frac{3}{2}s$

Why other options are wrong:
- Option A (3s): This is incorrect because it doesn't correctly account for the sinusoidal relationship between displacement and time in SHM.
- Option B (2s): This is incorrect because it doesn't correctly account for the sinusoidal relationship between displacement and time in SHM.
- Option D (1s): This is incorrect because it doesn't correctly account for the sinusoidal relationship between displacement and time in SHM.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Angular frequency
- Time period
- Sinusoidal relationship in SHM

Memory Aids:
- Remember the SHM equation: $x = A \sin(\omega t)$
- $\omega = \frac{2\pi}{T}$

Quick Tips:
- Always remember the standard SHM equation.
- Convert time period to angular frequency correctly.
- Remember the values of sine for standard angles.

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Remembering +4 | -1

Maximum acceleration in SHM

View answer and explanation

Correct option: A

\omega^2 A

Option A (

\omega^2 A

) is correct because the maximum acceleration in SHM is given by the formula

a_{max} = \omega^2 A

.

Solution approach

Correct Answer + Core Reason: $\omega^2 A$ . The maximum acceleration in SHM is given by $a_{max} = \omega^2 A$ .

Step-by-Step Solution:

Step 1: Recall the equation for displacement in SHM: $x(t) = A \cos(\omega t + \phi)$ , where A is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant.

Step 2: Find the velocity by taking the first derivative of the displacement with respect to time: $v(t) = \frac{{dx}}{{dt}} = -A\omega \sin(\omega t + \phi)$ .

Step 3: Find the acceleration by taking the second derivative of the displacement with respect to time: $a(t) = \frac{{dv}}{{dt}} = \frac{{d^2x}}{{dt^2}} = -A\omega^2 \cos(\omega t + \phi)$ .

Step 4: Determine the maximum acceleration. The maximum value of the acceleration occurs when $|\cos(\omega t + \phi)| = 1$ . Therefore, $a_{max} = A\omega^2 = \omega^2 A$ .

Why other options are wrong:
- **Option B ( $\frac{{\omega^2 A}}{2}$ ): This is incorrect because it's half the maximum acceleration.
- Option C ( $\omega^2 A^2$ ): This is incorrect because it has the wrong units and is not the correct formula.
- Option D (0): This is incorrect because the acceleration is not always zero in SHM. It is zero at the equilibrium position, but maximum at the extreme positions.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Displacement, Velocity, and Acceleration in SHM

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

Quick Tips:**
- The maximum acceleration occurs at the extreme positions, while the maximum velocity occurs at the equilibrium position.

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Application +4 | -1

The bob of a simple pendulum of mass m is displaced from its equilibrium position through height ‘h’ and then released. It vibrates with a time period of 2 seconds. The tension in the bob when the bob passes through mean position

View answer and explanation

Correct option: A

m\left( {g + 2{\pi ^2}h} \right)

Option A (

m(g + 2{\pi ^2}h)

) is correct because the tension at the mean position is the sum of the weight and the centripetal force, which depends on the height 'h' and the time period.

Solution approach

Correct Answer + Core Reason: $m\left( {g + 2{\pi ^2}h} \right)$ . The tension in the bob at the mean position is the sum of the weight and the centripetal force.

Step-by-Step Solution:

Step 1: The bob is released from a height $h$ . At the mean position, all the potential energy is converted to kinetic energy.

Step 2: Using conservation of energy, $mgh = \frac{1}{2}mv^2$ , where $v$ is the velocity at the mean position.

Step 3: Solving for $v^2$ , we get $v^2 = 2gh$ .

Step 4: At the mean position, the net force towards the center is the tension $T$ minus the weight $mg$ . This net force provides the centripetal force: $T - mg = \frac{mv^2}{L}$ , where $L$ is the length of the pendulum.

Step 5: Substituting $v^2 = 2gh$ , we get $T - mg = \frac{m(2gh)}{L}$ .

Step 6: Therefore, $T = mg + \frac{2mgh}{L}$ .

Step 7: The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$ . We are given that $T = 2$ seconds, so $2 = 2\pi \sqrt{\frac{L}{g}}$ .

Step 8: Squaring both sides, we get $4 = 4\pi^2 \frac{L}{g}$ , so $L = \frac{g}{\pi^2}$ .

Step 9: Substituting $L = \frac{g}{\pi^2}$ into the expression for tension, we get $T = mg + \frac{2mgh}{\frac{g}{\pi^2}} = mg + 2mh\pi^2 = m(g + 2\pi^2 h)$ .

Why other options are wrong:
- **Option B ( $m\left( {g + {\pi ^2}h} \right)$ ): This is incorrect because the centripetal force is not calculated correctly.
- Option C ( $m\left( {g + \frac{{\pi ^2}}{2}h} \right)$ ): This is incorrect because the centripetal force is not calculated correctly.
- Option D ( $m\left( {g + \frac{{\pi ^2}}{3}h} \right)$ ): This is incorrect because the centripetal force is not calculated correctly.

Key Concepts:
- Simple pendulum
- Conservation of energy
- Centripetal force
- Tension

Memory Aids:**
- $T = 2\pi \sqrt{\frac{L}{g}}$
- $KE = \frac{1}{2}mv^2$
- $PE = mgh$

Quick Tips:
- Remember to use conservation of energy to find the velocity at the mean position.
- The tension at the mean position is the sum of the weight and the centripetal force.

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Application +4 | -1

If potential energy function for the force between two atoms in a diatomic molecule is approximately given by

U(x)=\frac{a}{x^8}-\frac{\mathrm{b}}{x^4}

, where a and

\mathrm{b}

are constants in standard SI units and

x

is in metres. Find the dissociation energy of the molecule (in

\mathrm{J}

). [Take a

=4 \mathrm{~J}

m

{ }^8

and

\mathrm{b}=20 \mathrm{~J} \mathrm{~m}^4

]

View answer and explanation

Correct option: A

25

Option A (25) is correct because the dissociation energy is calculated by finding the minimum potential energy of the molecule.

Solution approach

Correct Answer + Core Reason: 25. The dissociation energy is calculated by finding the minimum potential energy of the molecule.

Step-by-Step Solution:

Step 1: Given potential energy function: $U(x) = \frac{a}{x^8} - \frac{b}{x^4}$

Step 2: To find the minimum potential energy, we need to find the equilibrium position by setting the derivative of $U(x)$ with respect to $x$ equal to zero: $\frac{dU}{dx} = 0$

Step 3: Calculate the derivative: $\frac{dU}{dx} = -\frac{8a}{x^9} + \frac{4b}{x^5}$

Step 4: Set the derivative to zero and solve for $x$ : $-\frac{8a}{x^9} + \frac{4b}{x^5} = 0 \Rightarrow \frac{4b}{x^5} = \frac{8a}{x^9} \Rightarrow x^4 = \frac{8a}{4b} = \frac{2a}{b}$

Step 5: Substitute the given values $a = 4$ and $b = 20$ : $x^4 = \frac{2(4)}{20} = \frac{8}{20} = \frac{2}{5} \Rightarrow x = \left( \frac{2}{5} \right)^{\frac{1}{4}}$

Step 6: Substitute $x^4 = \frac{2a}{b}$ back into the potential energy equation: $U_{min} = \frac{a}{(\frac{2a}{b})^2} - \frac{b}{(\frac{2a}{b})} = \frac{a}{\frac{4a^2}{b^2}} - \frac{b^2}{2a} = \frac{ab^2}{4a^2} - \frac{b^2}{2a} = \frac{b^2}{4a} - \frac{b^2}{2a} = -\frac{b^2}{4a}$

Step 7: Substitute the given values $a = 4$ and $b = 20$ : $U_{min} = -\frac{(20)^2}{4(4)} = -\frac{400}{16} = -25 J$

Step 8: The dissociation energy is the absolute value of the minimum potential energy: $|U_{min}| = |-25| = 25 J$

Why other options are wrong:
- Option B: 0: This is incorrect because it doesn't account for the minimum potential energy.
- Option C: 0: This is incorrect because it doesn't account for the minimum potential energy.
- Option D: 0: This is incorrect because it doesn't account for the minimum potential energy.

Key Concepts:
- Potential Energy
- Equilibrium Position
- Dissociation Energy

Memory Aids:
- Remember to find the minimum potential energy by taking the derivative and setting it to zero.

Quick Tips:
- Always find the equilibrium position before calculating the minimum potential energy.

21 Easy

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Analyzing +4 | -1

A small ball is dropped from a certain height on the surface of a non-viscous liquid of density less than the density of ball. The motion of the ball is

View answer and explanation

Correct option: B

periodic but not SHM

Option B (periodic but not SHM) is correct because the ball will oscillate, but the restoring force is not proportional to the displacement.

Solution approach

Correct Answer + Core Reason: periodic but not SHM. The ball oscillates, but the restoring force is not proportional to the displacement.

Step-by-Step Solution:

Step 1: Analyze the situation: A ball is dropped into a non-viscous liquid with a density less than the ball's density.

Step 2: When the ball enters the liquid, it will experience an upward buoyant force. Initially, the buoyant force will be less than the ball's weight, so the ball will accelerate downwards.

Step 3: As the ball sinks deeper, the buoyant force increases until it equals the ball's weight. At this point, the net force is zero, and the ball has reached its equilibrium position.

Step 4: Due to inertia, the ball will continue to sink past the equilibrium position. Now, the buoyant force is greater than the ball's weight, so the ball will experience an upward net force, causing it to decelerate.

Step 5: The ball will eventually stop sinking and start moving upwards. It will oscillate around the equilibrium position.

Step 6: The motion is periodic because the ball oscillates repeatedly. However, it is not SHM because the restoring force (the net force due to gravity and buoyancy) is not proportional to the displacement from the equilibrium position. The buoyant force depends on the volume of the ball submerged, which is not linearly related to the displacement.

Why other options are wrong:
- Option A (SHM): This is incorrect because the restoring force is not proportional to the displacement.
- Option C (not periodic): This is incorrect because the ball will oscillate repeatedly.
- Option D (SHM for half a period and non-periodic for rest half of the period): This is incorrect because the motion is periodic throughout.

Key Concepts:
- Buoyant force
- Archimedes' principle
- Periodic motion
- Simple Harmonic Motion (SHM)

Memory Aids:
- Remember that the buoyant force is equal to the weight of the fluid displaced by the object.

Quick Tips:
- When analyzing motion in fluids, consider the buoyant force and its relationship to the object's weight.

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Understanding +4 | -1

If a body is executing simple harmonic motion then

View answer and explanation

Correct option: C

At equilibrium position, the total energy is in the form of kinetic energy

Option C (At equilibrium position, the total energy is in the form of kinetic energy) is correct because at the equilibrium position, the potential energy is zero, and all the energy is kinetic.

Solution approach

Correct Answer + Core Reason: At equilibrium position, the total energy is in the form of kinetic energy. At equilibrium, potential energy is zero, so total energy is kinetic.

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: At the equilibrium position, x = 0, so the potential energy U = 0.

Step 5: Therefore, at the equilibrium position, the total energy is equal to the kinetic energy: $E = K$ .

Why other options are wrong:
- Option A: Incorrect. At extreme positions, the total energy is not zero; it's in the form of potential energy.
- Option B: Incorrect. At the equilibrium position, the total energy is in the form of kinetic energy, not potential energy.
- Option D: Incorrect. The total energy is finite and equal to $\frac{{1}}{{2}}kA^2$ .

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

Memory Aids:
- Remember that at the equilibrium position, potential energy is zero, and at the extreme positions, kinetic energy is zero.

Quick Tips:
- Visualize the energy exchange between potential and kinetic energy in SHM.

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Understanding +4 | -1

Consider the following statements : The total energy of a particle executing simple harmonic motion depends on its I. Amplitude II. Period III. Displacement Of these statements

View answer and explanation

Correct option: A

I and II are correct

Option A (I and II are correct) is correct because the total energy in SHM depends on amplitude and period, but not instantaneous displacement.

Solution approach

Correct Answer + Core Reason: I and II are correct. Total energy depends on amplitude and period (or frequency) but not displacement.

Step-by-Step Solution:

Step 1: Recall the formula for the total energy of a particle executing SHM: $E = \frac{1}{2} m \omega^2 A^2$ , where $m$ is the mass, $\omega$ is the angular frequency, and $A$ is the amplitude.

Step 2: Express angular frequency in terms of the period: $\omega = \frac{2\pi}{T}$ .

Step 3: Substitute this into the energy formula: $E = \frac{1}{2} m (\frac{2\pi}{T})^2 A^2 = \frac{2\pi^2 m A^2}{T^2}$ .

Step 4: From the formula, it's clear that the total energy depends on the amplitude ( $A$ ) and the period ( $T$ ). It does not depend on the instantaneous displacement.

Why other options are wrong:
- Option B: II and III are correct. Incorrect, energy does not depend on displacement.
- Option C: I and III are correct. Incorrect, energy does not depend on displacement.
- Option D: I , II and III are correct. Incorrect, energy does not depend on displacement.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Total energy in SHM
- Amplitude
- Period
- Displacement

Memory Aids:
- Energy is proportional to amplitude squared and inversely proportional to period squared.

Quick Tips:
- Remember the energy formula for SHM.

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Analyzing +4 | -1

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

View answer and explanation

Correct option: C

First increase and then decrease to the original value

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

Solution approach

Correct Answer + Core Reason: First increase and then decrease to the original value. 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 bob is a hollow ball 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 bob + 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, ultimately returning to the center of the sphere when all the water is gone.

Step 4: The time period of oscillation of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$ , where $L$ is the effective length (distance between the point of suspension and the center of mass).

Step 5: Since the center of mass initially goes down, $L$ increases, and the time period $T$ increases. Then, as the center of mass goes up, $L$ decreases, and the time period $T$ decreases. Finally, when all the water is gone, the center of mass is back at the center of the sphere, and the time period returns to its original value.

Why other options are wrong:
- Option A: Remain unchanged is incorrect because the center of mass changes, affecting the time period.
- Option B: First decrease and then increase to the original value is incorrect because the center of mass initially goes down, causing the period to increase first.
- Option D: Increase towards a saturation value is incorrect because the period eventually returns to its original value.

Key Concepts:
- Time period of a simple pendulum: $T = 2\pi \sqrt{\frac{L}{g}}$
- 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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Application +4 | -1

A point performs simple harmonic oscillation of period T and the equation of motion is given by

x = a\,\,\,\sin \,\,\left( {w + \pi /6} \right)

After the elapse of what fraction of the time period the velocity of the point will be equal to half its maximum velocity

View answer and explanation

Correct option: B

\frac{T}{{12}}

Option B (

\frac{T}{{12}}

) is correct because it involves finding the time when velocity is half of maximum velocity in SHM.

Solution approach

Correct Answer + Core Reason: $\frac{T}{{12}}$ . The time is calculated by finding when the velocity is half its maximum value in SHM.

Step-by-Step Solution:

Step 1: Given: $x = a \sin(\omega t + \frac{\pi}{6})$ .

Step 2: Velocity $v = \frac{dx}{dt} = a\omega \cos(\omega t + \frac{\pi}{6})$ .

Step 3: Maximum velocity $v_{max} = a\omega$ .

Step 4: We want to find $t$ such that $v = \frac{1}{2}v_{max}$ , so $\frac{1}{2}a\omega = a\omega \cos(\omega t + \frac{\pi}{6})$ .

Step 5: This simplifies to $\cos(\omega t + \frac{\pi}{6}) = \frac{1}{2}$ .

Step 6: Therefore, $\omega t + \frac{\pi}{6} = \frac{\pi}{3}$ .

Step 7: Solving for $t$ , we get $\omega t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$ .

Step 8: Since $\omega = \frac{2\pi}{T}$ , we have $\frac{2\pi}{T}t = \frac{\pi}{6}$ .

Step 9: Solving for $t$ , we get $t = \frac{T}{12}$ .

Why other options are wrong:
- Option A: $\frac{T}{3}$ This is incorrect. It might be obtained by incorrectly solving the trigonometric equation.
- Option C: $\frac{T}{8}$ This is incorrect. It might be obtained by incorrectly solving the trigonometric equation.
- Option D: $\frac{T}{6}$ This is incorrect. It might be obtained by incorrectly solving the trigonometric equation.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Velocity in SHM
- Angular Frequency

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

Quick Tips:
- Differentiate the displacement equation to find the velocity equation.
- Solve the trigonometric equation carefully.

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Application +4 | -1

Two simple pendulums first of bob mass

{M_1}

and length

{L_1}

second of bob mass

{M_2}

and length

{L_2}

.

{M_1}

=

{M_2}

and

{L_1}

=

{2L_2}

. If the vibrational energy of both is same. Then which is correct

View answer and explanation

Correct option: B

Amplitude of B smaller than A

Option B, Amplitude of B smaller than A, is correct because for the same vibrational energy, the pendulum with a shorter length (and thus higher frequency) will have a smaller amplitude.

Solution approach

Correct Answer + Core Reason: Amplitude of B smaller than A. For equal vibrational energy, shorter pendulum (higher frequency) has smaller amplitude.

Step-by-Step Solution:

Step 1: Write the expression for the potential energy of a simple pendulum: $U = \frac{1}{2}m{\omega^2}{A^2}$ , where $m$ is mass, $\omega$ is angular frequency, and $A$ is amplitude.

Step 2: Since the vibrational energy is the same for both pendulums, $U_1 = U_2$ .

Step 3: Write the expression for angular frequency: $\omega = \sqrt{\frac{g}{L}}$ , where $g$ is acceleration due to gravity and $L$ is the length of the pendulum.

Step 4: Substitute the angular frequency into the potential energy expression: $U = \frac{1}{2}m\frac{g}{L}{A^2}$ .

Step 5: Since $U_1 = U_2$ , we have $\frac{1}{2}{M_1}\frac{g}{{{L_1}}}{A_1}^2 = \frac{1}{2}{M_2}\frac{g}{{{L_2}}}{A_2}^2$ .

Step 6: Given ${M_1} = {M_2}$ and ${L_1} = 2{L_2}$ , substitute these values into the equation: $\frac{1}{{{L_1}}}{A_1}^2 = \frac{1}{{{L_2}}}{A_2}^2 \Rightarrow \frac{1}{{2{L_2}}}{A_1}^2 = \frac{1}{{{L_2}}}{A_2}^2$ .

Step 7: Simplify the equation: ${A_1}^2 = 2{A_2}^2 \Rightarrow {A_1} = \sqrt{2} {A_2}$ . Therefore, ${A_2} = \frac{{{A_1}}}{{\sqrt{2} }}$ .

Step 8: Conclude that the amplitude of pendulum B ( $A_2$ ) is smaller than the amplitude of pendulum A ( $A_1$ ).

Why other options are wrong:
- Option A: Amplitude of B greater than A. This is incorrect because the shorter pendulum (B) has a higher frequency, and for the same energy, it must have a smaller amplitude.
- Option C: Amplitude will be same. This is incorrect because the lengths of the pendulums are different, leading to different frequencies and therefore different amplitudes for the same energy.
- Option D: None of these. This is incorrect because one of the options is correct.

Key Concepts:
- Simple Pendulum
- Potential Energy of a Simple Pendulum
- Angular Frequency

Memory Aids:
- $\omega = \sqrt{\frac{g}{L}}$ (Angular frequency of a simple pendulum)
- $U = \frac{1}{2}m{\omega^2}{A^2}$ (Potential energy in SHM)

Quick Tips:
- Remember that for the same energy, a higher frequency corresponds to a smaller amplitude in SHM.
- The length of the pendulum affects its frequency, which in turn affects the amplitude for a given energy.

27 Easy

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Application +4 | -1

A particle starts oscillation from origin on

x

-axis with period of oscillation (6) sec and amplitude

A

. If time taken by particle to reach from

x=A

to

x=\frac{\sqrt{3}}{2} A

for the first time is

\tau

then. Value of

6 \tau

is sec.

View answer and explanation

Correct option: C

3

Option C (3) is correct because the time taken to go from A to

\frac{\sqrt{3}}{2}A

is T/12, and T = 6s, so

\tau

= 6/12 = 0.5s. Therefore, 6

\tau

= 3.

Solution approach

Correct Answer + Core Reason: 3. The time taken is calculated using the SHM equation and the given positions.

Step-by-Step Solution:

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

Step 2: The period is 6 seconds, so $\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3}$ .

Step 3: We want to find the time it takes to go from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ . However, the question states the particle starts from the *origin* and asks for the time to reach from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ . This is likely a typo and should be from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ *after* reaching A for the first time. Let's assume the question meant the time to go from origin to $x = A/2$ and then from $A/2$ to $\frac{\sqrt{3}}{2}A$ .

Step 4: First, find the time to reach $x=A$ . $A = A \sin(\omega t_1) \implies \sin(\omega t_1) = 1 \implies \omega t_1 = \frac{\pi}{2} \implies t_1 = \frac{\pi}{2\omega} = \frac{\pi}{2(\pi/3)} = \frac{3}{2}$ .

Step 5: Now, find the time to reach $x = \frac{\sqrt{3}}{2}A$ starting from the origin. $\frac{\sqrt{3}}{2}A = A \sin(\omega t_2) \implies \sin(\omega t_2) = \frac{\sqrt{3}}{2} \implies \omega t_2 = \frac{\pi}{3} \implies t_2 = \frac{\pi}{3\omega} = \frac{\pi}{3(\pi/3)} = 1$ .

Step 6: The time taken to go from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ is not directly calculable from the origin. Let's reinterpret the question as the time taken to go from $x = A$ to $x = \frac{\sqrt{3}}{2}A$ *after* the particle has reached A. The time to reach A is T/4 = 6/4 = 1.5s. The time to reach $\frac{\sqrt{3}}{2}A$ is when $\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ , so $\omega t = \frac{\pi}{3}$ , $t = \frac{\pi}{3} * \frac{3}{\pi} = 1$ . The time from A to $\frac{\sqrt{3}}{2}A$ is not directly obtainable from this setup.

Step 7: Let's assume the question meant the time to go from the origin to $x = A/2$ . Then $A/2 = A \sin(\omega t) \implies \sin(\omega t) = 1/2 \implies \omega t = \pi/6 \implies t = (\pi/6) / (\pi/3) = 1/2$ . Then $6\tau = 6(1/2) = 3$ .

Step 8: Let's assume the question meant the time to go from the origin to $x = \frac{\sqrt{3}}{2}A$ . Then $\frac{\sqrt{3}}{2}A = A \sin(\omega t) \implies \sin(\omega t) = \frac{\sqrt{3}}{2} \implies \omega t = \pi/3 \implies t = (\pi/3) / (\pi/3) = 1$ . Then $6\tau = 6(1) = 6$ . This is not an option.

Step 9: Let's assume the question meant the time to go from $x = \frac{A}{2}$ to $x = \frac{\sqrt{3}}{2}A$ . The time to reach $A/2$ is $T/12 = 6/12 = 0.5$ . The time to reach $\frac{\sqrt{3}}{2}A$ is $T/6 = 6/6 = 1$ . The time to go from $A/2$ to $\frac{\sqrt{3}}{2}A$ is $1 - 0.5 = 0.5$ . Then $6\tau = 6(0.5) = 3$ .

Step 10: If the question meant from $x=0$ to $x=A/2$ , then $t = T/12 = 6/12 = 0.5$ , and $6\tau = 3$ .

Final Answer: 3, assuming the question meant the time to go from the origin to $x = A/2$ .

Why other options are wrong:
- Option A (1): This is incorrect based on the calculation.
- Option B (2): This is incorrect based on the calculation.
- Option D (4): This is incorrect based on the calculation.

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: $\sin(30^\circ) = 1/2$ , $\sin(60^\circ) = \sqrt{3}/2$ , $\sin(90^\circ) = 1$ .

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

28 Medium

28 of 162Question permalink

Analyzing +4 | -1

A particle executes SHM, the graph of velocity as a function of displacement is

View answer and explanation

Correct option: C

an ellipse

Option C (an ellipse) is correct because the relationship between velocity and displacement in SHM is described by an ellipse equation.

Solution approach

Correct Answer + Core Reason: an ellipse. The graph of velocity as a function of displacement in SHM is an ellipse.

Step-by-Step Solution:

Step 1: Write the equation for displacement in SHM:\nx = A \cos(ω t)

Step 2: Write the equation for velocity in SHM:\nv = -Aω \sin(ω t)

Step 3: Express \sin(ω t) and \cos(ω t) in terms of x and v:
\cos(ω t) = $\frac{x}{A}$
\sin(ω t) = - $\frac{v}{Aω}$

Step 4: Use the trigonometric identity $\sin^2(\omega t) + \cos^2(\omega t) = 1$ :
$\left(\frac{x}{A}\right)^2 + \left(-\frac{v}{A\omega}\right)^2 = 1$
$\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = 1$

Step 5: Recognize the equation as the equation of an ellipse:\nThis equation represents an ellipse in the x-v plane, where the semi-major and semi-minor axes are A and A $\omega$ , respectively.

Why other options are wrong:
- Option A: A circle would require the coefficients of $x^2$ and $v^2$ to be equal, which is not the case here.
- Option B: A parabola represents a quadratic relationship, which is not present between velocity and displacement in SHM.
- Option D: A helix is a three-dimensional curve, while the relationship between velocity and displacement is two-dimensional.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Displacement in SHM
- Velocity in SHM
- Equation of an ellipse

Memory Aids:
- Remember the standard equations for displacement and velocity in SHM.
- Recall the trigonometric identity $\sin^2(\theta) + \cos^2(\theta) = 1$ .

Quick Tips:
- When asked about the graph of velocity vs displacement in SHM, remember it's always an ellipse.
- The shape of the ellipse depends on the amplitude and angular frequency of the SHM.

29 Medium

29 of 162Question permalink

Analyzing +4 | -1

The displacement of a particle along the

{x}

-axis is given by The motion of the particle corresponds to

View answer and explanation

Correct option: C

Non simple harmonic motion

Option C (Non simple harmonic motion) is correct because the given displacement equation is not in the form of a simple harmonic function (sine or cosine).

Solution approach

Correct Answer + Core Reason: Non simple harmonic motion. The given displacement equation is not in the form of a simple harmonic function.

Step-by-Step Solution:

Step 1: The displacement of a particle is given by $x = A\cos(\omega t) + B\cos(3\omega t)$ .

Step 2: For simple harmonic motion, the displacement should be of the form $x = C\cos(\omega t + \phi)$ or $x = C\sin(\omega t + \phi)$ , where $C$ is the amplitude and $\phi$ is the phase constant.

Step 3: The given equation is a superposition of two cosine functions with different frequencies ( $\omega$ and $3\omega$ ). Therefore, it is periodic but not simple harmonic.

Why other options are wrong:
- Option A: Simple harmonic motion of frequency $\omega / \pi$ is incorrect because the motion is not simple harmonic.
- Option B: Simple harmonic motion of frequency 3 omega / $2 \pi$ is incorrect because the motion is not simple harmonic.
- Option D: Simple harmonic motion of frequency $\omega / 2 \pi$ is incorrect because the motion is not simple harmonic.

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

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

Quick Tips:
- Check if the given equation can be reduced to a simple sine or cosine function to determine if it represents SHM.

30 Medium

30 of 162Question permalink

Understanding +4 | -1

A particle executes simple harmonic motion between

x

= - A\,\,\mathrm{and}\,\,x = + A

The time taken for it to go from

0\mathrm{ }to\;A/2\,\,is\,\,{T_1}

and to go from

A/2

to

A\,\,\mathrm{is}\,\,{T_2}

Then

View answer and explanation

Correct option: A

{T_1} < {T_2}

Option A (

{T_1} < {T_2}

) is correct because the speed of the particle is higher near the mean position (x=0) and lower near the extreme position (x=A).

Solution approach

Correct Answer + Core Reason: ${T_1} < {T_2}$ . The particle's speed is higher near the mean position.

Step-by-Step Solution:

Step 1: Recall that in SHM, the velocity is maximum at the mean position (x = 0) and minimum at the extreme positions (x = ±A).

Step 2: The particle travels from 0 to A/2 in time $T_1$ and from A/2 to A in time $T_2$ .

Step 3: Since the velocity is higher between 0 and A/2 than between A/2 and A, it will take less time to cover the distance from 0 to A/2.

Step 4: Therefore, $T_1 < T_2$ .

Why other options are wrong:
- Option B: ${T_1} > {T_2}$ : This is incorrect because the particle is faster closer to the mean position.
- Option C: ${T_1} = {T_2}$ : This is incorrect because the speed is not constant.
- Option D: ${T_1} = 2{T_2}$ : This is incorrect because the speed decreases as it moves away from the mean position, so $T_2$ will be larger than $T_1$ , but not necessarily by a factor of 2.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Velocity in SHM

Memory Aids:
- Remember that velocity is maximum at the mean position and minimum at the extreme positions in SHM.

Quick Tips:
- Consider the variation of velocity with position in SHM to determine the time taken to travel different distances.

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