The bob of a simple pendulum of length L is released at time $t{\rm{ }} = {\rm{ }}0$ from a position of small angular displacement. Its linear displacement at time t is given by

${\rm{X}} = {\rm{a cos}}\sqrt {\frac{{\rm{g}}}{{\rm{L}}}} \times {\rm{t}}$

$X = a\sin 2\pi \sqrt {\frac{L}{{\rm{g}}}} \times t$

$X = a\cos 2\pi \sqrt {\frac{{\rm{g}}}{L}} \times t$

${\rm{X}} = {\rm{a sin}}\sqrt {\frac{{\rm{g}}}{{\rm{L}}}} \times {\rm{t}}$

OSCILLATIONS MCQs for NEET UG PHYSICS — Medium, Page 3

61 Medium

Understanding +4 | -1

Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion

View answer and explanation

Correct option: D

When v is maximum, a is zero

Option D (When v is maximum, a is zero) is correct because in SHM, the acceleration is proportional to the displacement from the mean position, and the velocity is maximum at the mean position where displacement and hence acceleration is zero.

Solution approach

Correct Answer + Core Reason: When v is maximum, a is zero. In SHM, maximum velocity occurs at the equilibrium position where acceleration is zero.

Step-by-Step Solution:

Step 1: Recall the equations for velocity and acceleration in SHM:
- Velocity: $v = \omega \sqrt{A^2 - x^2}$
- Acceleration: $a = -\omega^2 x$

Step 2: Analyze the velocity equation. Velocity is maximum when $x = 0$ (at the mean position), and the maximum velocity is $v_{max} = \omega A$ .

Step 3: Analyze the acceleration equation. Acceleration is zero when $x = 0$ (at the mean position).

Step 4: Therefore, when velocity is maximum (at the mean position), acceleration is zero.

Why other options are wrong:
- Option A: When v is maximum, a is maximum: This is incorrect because when velocity is maximum, displacement is zero, and therefore acceleration is zero.
- Option B: Value of a is zero, whatever may be the value of v: This is incorrect because acceleration depends on displacement, which affects velocity.
- Option C: When v is zero, a is zero: This is incorrect because when velocity is zero (at the extreme positions), displacement is maximum, and therefore acceleration is maximum.

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

Memory Aids:
- Remember that in SHM, velocity and acceleration are out of phase.

Quick Tips:
- Remember the relationship between displacement, velocity, and acceleration in SHM. Velocity is maximum at the mean position, and acceleration is maximum at the extreme positions.

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Assertion (A) : If the elevator goes down with constant acceleration, then frequency of oscillation of spring block system in elevator will be independent of acceleration of elevator.
Reason (R) : If the given elevator moves with uniform velocity, then the frequency of oscillation will change.

View answer and explanation

Correct option: C

A is true, R is false

The assertion is true because the frequency of a spring-block system depends only on the spring constant and mass, not the elevator's acceleration. The reason is false because uniform velocity doesn't affect the frequency.

Solution approach

Correct Answer + Core Reason: A is true, R is false. The frequency of oscillation of a spring-block system is independent of the elevator's acceleration, and uniform velocity also doesn't change the frequency.

Step-by-Step Solution:

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

Step 2: Analyze the effect of constant acceleration on the system. The acceleration of the elevator only affects the equilibrium position of the mass but not the frequency of oscillation.

Step 3: Analyze the effect of uniform velocity on the system. Uniform velocity has no effect on the frequency of oscillation.

Step 4: Conclude that the assertion is true because the frequency is independent of the elevator's acceleration. The reason is false because uniform velocity does not change the frequency.

Why other options are wrong:
- Option A: A and R are true, R is correct explanation of A: R is false.
- Option B: A and R are true, R is not correct explanation of A: R is false.
- Option D: A is false, R is true: A is true and R is false.

Key Concepts:
- Frequency of oscillation of a spring-block system
- Effect of acceleration and velocity on SHM

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

Quick Tips:
- Remember the formula for the frequency of a spring-block system.

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Analyzing +4 | -1

Statement-I: If a pendulum clock is taken to mountain top, it gains time.
Statement-II: Value of acceleration due to gravity is less at heights.

View answer and explanation

Correct option: D

Statement-I is false

\&

Statement-II is true

Option D (Statement-I is false & Statement-II is true) is correct because a pendulum clock loses time on a mountain top due to reduced gravity, and the value of g is indeed less at higher altitudes.

Solution approach

Correct Answer + Core Reason: Statement-I is false $\& Statement-II is true. A pendulum clock loses time on a mountain top, and the value of g is less at heights.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Statement I: If a pendulum clock is taken to mountain top, it gains time. This is false. The time period of a pendulum is given by$ T = 2π $\sqrt{\frac{L}{g}}$ $. If$ g $decreases,$ T $increases, meaning the clock runs slower and loses time.\n\n**Step 2:** Statement II: Value of acceleration due to gravity is less at heights. This is true. The acceleration due to gravity decreases with height according to the formula$ g' = g(1 - $\frac{2h}{R}$ ) $, where$ h $is the height and$ R $is the radius of the Earth.\n\n**Why other options are wrong:**\n- **Option A:** Statement-I$ \& Statement-II are true. This is incorrect because Statement I is false.
- Option B: Statement-I $\& Statement-II are false. This is incorrect because Statement II is true.\n- **Option C:** Statement-I is true$ \& Statement-II is false. This is incorrect because Statement I is false and Statement II is true.

Key Concepts:
- Simple Pendulum
- Time Period
- Acceleration due to Gravity
- Variation of g with Height

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

Quick Tips:
- Understand how the time period of a pendulum depends on the acceleration due to gravity.
- Know that the acceleration due to gravity decreases with height.

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In S.H.M. which one of the following quantities has constant ratio with acceleration:

View answer and explanation

Correct option: B

Displacement

Option B (Displacement) is correct because in SHM, acceleration is directly proportional to the negative of the displacement, making their ratio constant.

Solution approach

Correct Answer + Core Reason: Displacement. In SHM, acceleration is proportional to the negative of displacement.

Step-by-Step Solution:

Step 1: Recall the equation for acceleration in SHM: $a = -\omega^2 x$ , where $a$ is the acceleration, $\omega$ is the angular frequency, and $x$ is the displacement from the equilibrium position.

Step 2: Analyze the ratio of acceleration to displacement: $\frac{a}{x} = -\omega^2$ . Since $\omega$ is constant for a given SHM, the ratio of acceleration to displacement is constant.

Why other options are wrong:
- Option A (Time): This is incorrect because the acceleration is not directly proportional to time.
- Option C (Velocity): This is incorrect because the acceleration is

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Sine and cosine functions are:

View answer and explanation

Correct option: C

Periodic functions

Option C (Periodic functions) is correct because sine and cosine functions repeat their values after a fixed interval, making them periodic.

Solution approach

Correct Answer + Core Reason: Periodic functions. Sine and cosine functions repeat their values after a fixed interval.

Step-by-Step Solution:

Step 1: Definition of a periodic function: A function f(x) is periodic if there exists a non-zero constant T such that $f(x + T) = f(x)$ for all x.

Step 2: Sine function: $\sin(x + 2\pi) = \sin(x)$ .

Step 3: Cosine function: $\cos(x + 2\pi) = \cos(x)$ .

Step 4: Conclusion: Since sine and cosine functions satisfy the definition of a periodic function, they are periodic functions.

Why other options are wrong:
- Option A: Incorrect. Algebraic functions are polynomials or rational functions.
- Option B: Incorrect. Logarithmic functions are of the form $y = \log_b(x)$ .
- Option D: Incorrect. Sine and cosine functions are periodic.

Key Concepts:
- Periodic Functions
- Sine Function
- Cosine Function

Memory Aids:
- Remember the graphs of sine and cosine functions and their repeating nature.

Quick Tips:
- Recognize the periodic nature of trigonometric functions.

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Application +4 | -1

The displacement of a particle in SHM various according to the relation

x = 4\left( {\mathrm{cos\pi }t + \mathrm{sin\pi }t} \right)

The amplitude of the particle is

View answer and explanation

Correct option: C

4\sqrt 2

Option C (

4\sqrt{2}

) is correct because the given equation can be rewritten as

x = 4\sqrt{2} \sin(\pi t + \frac{\pi}{4})

, so the amplitude is

4\sqrt{2}

.

Solution approach

Correct Answer + Core Reason: $4\sqrt{2}$ . The amplitude is found by rewriting the equation in standard SHM form.

Step-by-Step Solution:

Step 1: Given the equation $x = 4(\cos \pi t + \sin \pi t)$ .

Step 2: Rewrite the equation in the form $x = A \sin(\pi t + \phi)$ .

Step 3: We can write $x = A(\sin \pi t \cos \phi + \cos \pi t \sin \phi)$ .

Step 4: Comparing coefficients, we have $A \cos \phi = 4$ and $A \sin \phi = 4$ .

Step 5: Squaring and adding these equations, we get $A^2(\cos^2 \phi + \sin^2 \phi) = 4^2 + 4^2$ .

Step 6: Simplifying, $A^2 = 16 + 16 = 32$ .

Step 7: Therefore, $A = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$ .

Why other options are wrong:
- **Option A ( $-4$ ): This is incorrect because amplitude is always a positive quantity.
- Option B ( $4$ ): This is incorrect because it only considers the coefficient of either the sine or cosine term, not the combined effect.
- Option D ( $8$ ): This is incorrect because it's the sum of the coefficients, not the amplitude of the combined SHM.

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

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

Quick Tips:
- When dealing with a sum of sine and cosine functions with the same frequency, rewrite it as a single sine or cosine function with a phase shift.

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Application +4 | -1

The angular frequency of a simple pendulum is

\omega \,rad/s

. Now the length is made one fourth of the original length , the angular frequency becomes

View answer and explanation

Correct option: B

2\omega

Option B (

2\omega

) is correct because the angular frequency is inversely proportional to the square root of the length, so reducing the length to one-fourth doubles the angular frequency.

Solution approach

Correct Answer + Core Reason: $2\omega$ . When the length is made one fourth of the original length, the angular frequency becomes twice the original value.

Step-by-Step Solution:

Step 1: The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g}{L}}$ , where $g$ is the acceleration due to gravity and $L$ is the length of the pendulum.

Step 2: Let the original length be $L_1 = L$ and the new length be $L_2 = \frac{L}{4}$ .

Step 3: The original angular frequency is $\omega_1 = \sqrt{\frac{g}{L}}$ .

Step 4: The new angular frequency is $\omega_2 = \sqrt{\frac{g}{L_2}} = \sqrt{\frac{g}{\frac{L}{4}}} = \sqrt{\frac{4g}{L}} = 2\sqrt{\frac{g}{L}}$ .

Step 5: Therefore, $\omega_2 = 2\omega_1 = 2\omega$ .

Why other options are wrong:
- **Option A ( $\frac{\omega }{2}$ ): This is incorrect because the angular frequency increases when the length decreases.
- Option C ( $4\omega$ ): This is incorrect because the angular frequency is proportional to the square root of the inverse of the length, not the inverse itself.
- Option D ( $\omega$ ): This is incorrect because the angular frequency changes when the length changes.

Key Concepts:
- Simple pendulum
- Angular frequency

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

Quick Tips:
- Remember the relationship between angular frequency and length for a simple pendulum. If the length is reduced by a factor of $n$ , the angular frequency increases by a factor of $\sqrt{n}$ .

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Understanding +4 | -1

Displacement between maximum potential energy position and maximum kinetic energy position for a particle executing S.H.M. is

View answer and explanation

Correct option: C

\pm a

Option C (

\pm a

) is correct because the maximum potential energy position is at the extreme end (

\pm a

) and the maximum kinetic energy position is at the mean position (0), so the displacement is

\pm a

.

Solution approach

Correct Answer + Core Reason: $\pm a$ . The displacement between maximum potential energy and maximum kinetic energy positions is the amplitude.

Step-by-Step Solution:

Step 1: Maximum potential energy position: In SHM, the potential energy is maximum at the extreme positions, which are $x = +a$ and $x = -a$ , where a is the amplitude.

Step 2: Maximum kinetic energy position: In SHM, the kinetic energy is maximum at the equilibrium position, which is $x = 0$ .

Step 3: Displacement between maximum potential energy and maximum kinetic energy positions: The displacement is either $+a - 0 = +a$ or $-a - 0 = -a$ .

Step 4: Conclusion: The displacement is $\pm a$ .

Why other options are wrong:
- Option A: Incorrect. The displacement is not just -a.
- Option B: Incorrect. The displacement is not just +a.
- Option D: Incorrect. The displacement is not $\pm a/4$ .

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

Memory Aids:
- Remember that potential energy is maximum at the extremes, and kinetic energy is maximum at the equilibrium position.

Quick Tips:
- Visualize the positions of maximum potential and kinetic energy in SHM.

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Application +4 | -1

The equation of SHM is given by

x = 3\sin 20\pi t + 4\cos 20\pi t

Where x is in cm and t is I second. The amplitude is

View answer and explanation

Correct option: C

5 cm

Option C (5 cm) is correct because the amplitude is calculated by finding the resultant of the two SHM components.

Solution approach

Correct Answer + Core Reason: 5 cm. The amplitude is the square root of the sum of the squares of the coefficients of sine and cosine.

Step-by-Step Solution:

Step 1: Identify the given equation: $x = 3\sin 20\pi t + 4\cos 20\pi t$ .

Step 2: Recognize that this is a superposition of two SHMs with the same frequency but different phases.

Step 3: The amplitude of the resultant 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 $A_1 = 3$ cm and $A_2 = 4$ cm.

Step 5: Calculate the resultant amplitude: $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{{ cm}}$ .

Why other options are wrong:
- Option A (7 cm): This is incorrect because it simply adds the two amplitudes instead of using the Pythagorean theorem to find the resultant amplitude.
- Option B (4 cm): This is incorrect because it only considers the amplitude of the cosine term and ignores the sine term.
- Option D (3 cm): This is incorrect because it only considers the amplitude of the sine term and ignores the cosine term.

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.

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Understanding +4 | -1

To show that a simple pendulum executes simple harmonic motion, it is necessary to assume that

View answer and explanation

Correct option: C

Amplitude of oscillation is small

Option C (Amplitude of oscillation is small) is correct because the approximation

\sin \theta \approx \theta

is valid only for small angles, which is necessary for SHM.

Solution approach

Correct Answer + Core Reason: Amplitude of oscillation is small. For SHM, $\sin \theta \approx \theta$ must hold, which is true for small amplitudes.

Step-by-Step Solution:

Step 1: Recall the equation of motion for a simple pendulum: $\frac{d^2\theta}{dt^2} + \frac{g}{L} \sin \theta = 0$ .

Step 2: For small oscillations, we can approximate $\sin \theta \approx \theta$ . This simplifies the equation to $\frac{d^2\theta}{dt^2} + \frac{g}{L} \theta = 0$ , which is the equation for SHM.

Step 3: If the amplitude is not small, the approximation $\sin \theta \approx \theta$ is not valid, and the motion is not simple harmonic.

Why other options are wrong:
- Option A: Length of the pendulum is small: The length of the pendulum affects the time period but not whether the motion is SHM.
- Option B: Mass of the pendulum is small: The mass of the pendulum does not affect whether the motion is SHM.
- Option D: Acceleration due to gravity is small: The acceleration due to gravity affects the time period but not whether the motion is SHM.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Simple pendulum
- Small angle approximation

Memory Aids:
- $\sin \theta \approx \theta$ for small $\theta$ .

Quick Tips:
- Remember the small angle approximation is crucial for SHM in a simple pendulum.

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Analyzing +4 | -1

A particle moves with simple harmonic motion in a straight line. In first

\tau

sec, after starting from rest it travels a distance

a

and in next

\tau

sec, it travels

2a

, in same direction, then

View answer and explanation

Correct option: D

time period of oscillations is

6\pi

Option D (time period of oscillations is

6\tau

) is the correct answer. The problem describes a particle starting from rest and moving in SHM. By analyzing the distances traveled in consecutive time intervals, we can determine the time period.

Solution approach

Correct Answer + Core Reason: time period of oscillations is $6\tau$ . By analyzing the distances traveled in consecutive time intervals, we can determine the time period.

Step-by-Step Solution:

Step 1: Let the amplitude be $A$ . Since the particle starts from rest, we can write the displacement as $x = A \cos(\omega t)$ .

Step 2: At $t = \tau$ , the displacement is $A - a$ . So, $A - a = A \cos(\omega \tau)$ .

Step 3: At $t = 2\tau$ , the displacement is $A - a - 2a = A - 3a$ . So, $A - 3a = A \cos(2 \omega \tau)$ .

Step 4: Let $\theta = \omega \tau$ . Then, we have $A - a = A \cos(\theta)$ and $A - 3a = A \cos(2\theta)$ .

Step 5: From the first equation, $\cos(\theta) = 1 - \frac{a}{A}$ . From the second equation, $\cos(2\theta) = 1 - \frac{3a}{A}$ .

Step 6: Using the identity $\cos(2\theta) = 2 \cos^2(\theta) - 1$ , we have $1 - \frac{3a}{A} = 2(1 - \frac{a}{A})^2 - 1$ .

Step 7: Simplifying, $1 - \frac{3a}{A} = 2(1 - \frac{{2a}}{A} + \frac{{a^2}}{{A^2}}) - 1 \Rightarrow 1 - \frac{3a}{A} = 2 - \frac{{4a}}{A} + \frac{{2a^2}}{{A^2}} - 1 \Rightarrow 0 = - \frac{a}{A} + \frac{{2a^2}}{{A^2}}$ .

Step 8: Multiplying by $A^2$ , $0 = -aA + 2a^2 \Rightarrow A = 2a$ .

Step 9: Substituting $A = 2a$ into $\cos(\theta) = 1 - \frac{a}{A}$ , we get $\cos(\theta) = 1 - \frac{a}{{2a}} = 1 - \frac{1}{2} = \frac{1}{2}$ .

Step 10: Therefore, $\theta = \frac{\pi}{3}$ . Since $\theta = \omega \tau$ , we have $\omega \tau = \frac{\pi}{3}$ .

Step 11: The time period $T = \frac{{2\pi}}{\omega}$ . So, $\omega = \frac{{2\pi}}{T}$ . Substituting this into $\omega \tau = \frac{\pi}{3}$ , we get $\frac{{2\pi}}{T} \tau = \frac{\pi}{3} \Rightarrow T = 6\tau$ .

Why other options are wrong:
- **Option A (amplitude of motion is $3a$ ): The amplitude is 2a, not 3a.
- Option B (time period of oscillations is $8\tau$ ):** The time period is $6\tau$ , not $8\tau$ .
- **Option C (amplitude of motion is $4a$ ): The amplitude is 2a, not 4a.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Displacement in SHM
- Time period of SHM

Memory Aids:**
- Remember the displacement equation for SHM starting from rest: $x = A \cos(\omega t)$ .

Quick Tips:
- Use the given information to set up equations and solve for the unknowns (amplitude and time period).

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Remembering +4 | -1

x(t)=A \cos (\omega t+\phi)

is the equation of simple harmonic motion. In this equation '

\phi

' is called

View answer and explanation

Correct option: A

phase constant

Option A (phase constant) is correct because '

\phi

' in the equation

x(t)=A \cos (\omega t+\phi)

represents the initial phase or phase constant.

Solution approach

Correct Answer + Core Reason: phase constant. In the equation of SHM, $\phi$ represents the phase constant.

Step-by-Step Solution:

Step 1: General equation of SHM: $x(t) = A \cos(\omega t + \phi)$ .

Step 2: Identify the terms:
- A: Amplitude
- $\omega$ : Angular frequency
- t: Time
- $\phi$ : Phase constant (initial phase).

Step 3: Conclusion: $\phi$ is the phase constant.

Why other options are wrong:
- Option B: Incorrect. Frequency is represented by $\omega$ (angular frequency) or f (linear frequency).
- Option C: Incorrect. Amplitude is represented by A.
- Option D: Incorrect. Displacement is represented by x(t).

Key Concepts:
- Simple Harmonic Motion (SHM)
- Phase Constant
- Amplitude
- Frequency

Memory Aids:
- Remember the general form of the SHM equation and the meaning of each term.

Quick Tips:
- Memorize the standard equation for SHM and identify each parameter.

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Analyzing +4 | -1

A body is moving in a room with a velocity of 20 m/s perpendicular to the two walls separated by 5 meters. There is no friction and the collisions with the walls are elastic. The motion of the body is

View answer and explanation

Correct option: B

Periodic but not simple harmonic

Option B (Periodic but not simple harmonic) is correct because the motion repeats at regular intervals, but the acceleration is not proportional to the displacement from a fixed point.

Solution approach

Correct Answer + Core Reason: Periodic but not simple harmonic. The motion repeats at regular intervals, but the acceleration is not proportional to the displacement.

Step-by-Step Solution:

Step 1: The body moves with a constant velocity of 20 m/s perpendicular to the walls separated by 5 meters. The collisions with the walls are elastic, meaning the velocity changes direction but the speed remains the same.

Step 2: The time taken to travel from one wall to the other is $t = \frac{d}{v} = \frac{5 \text{ m}}{20 \text{ m/s}} = 0.25 \text{ s}$ .

Step 3: The body travels from one wall to the other and back, so the total time for one complete cycle is $T = 2t = 2 \times 0.25 \text{ s} = 0.5 \text{ s}$ . Since the motion repeats after a fixed interval of time, it is periodic.

Step 4: For simple harmonic motion, the acceleration must be proportional to the displacement from a fixed point and directed towards that point. In this case, the acceleration is zero between the walls, and during the collision, it is very large but not proportional to displacement. Therefore, the motion is not simple harmonic.

Why other options are wrong:
- Option A: Not periodic is incorrect because the motion repeats after a fixed interval of time.
- Option C: Periodic and simple harmonic is incorrect because the acceleration is not proportional to the displacement.
- Option D: Periodic with variable time period is incorrect because the time period is constant.

Key Concepts:
- Periodic motion
- Simple harmonic motion (SHM)

Memory Aids:
- Remember the conditions for SHM: acceleration proportional to displacement and directed towards the mean position.

Quick Tips:
- Check if the acceleration is proportional to the displacement to determine if the motion is SHM.

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Analyzing +4 | -1

The displacement of a particle varies according to the relation

x=3 \sin 100 t+8 \cos ^2 50 t

. Which of the following is incorrect about this motion.

View answer and explanation

Correct option: C

The amplitude of the resultant S.H. M. is

\sqrt{73}

units

Option C (The amplitude of the resultant S.H. M. is

\sqrt{73}

units) is incorrect because the expression can be simplified to a sum of two SHM terms, and the amplitude is not simply

\sqrt{73}

.

Solution approach

Correct Answer + Core Reason: The amplitude of the resultant S.H. M. is $\sqrt{73}$ units. This statement is incorrect because the given equation is not a simple SHM, but a combination of SHMs, and the amplitude is not simply calculated as $\sqrt{73}$ .

Step-by-Step Solution:

Step 1: Rewrite the given equation using the trigonometric identity $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ : $x = 3 \sin(100t) + 8 \cos^2(50t) = 3 \sin(100t) + 8 \left( \frac{1 + \cos(100t)}{2} \right) = 3 \sin(100t) + 4 + 4 \cos(100t)$ .

Step 2: Combine the sine and cosine terms into a single SHM: $x = 4 + 3 \sin(100t) + 4 \cos(100t)$ . The amplitude of the SHM part is $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ .\nSo, $x = 4 + 5 \sin(100t + \phi)$ , where $\phi = \arctan(\frac{4}{3})$ .

Step 3: Analyze the statements:
- A: The motion of the particle is S.H.M. - Incorrect, it is SHM with a constant displacement of 4 units.
- B: The amplitude of the S.H.M. of the particle is 5 units - Correct, as calculated above.
- C: The amplitude of the resultant S.H. M. is $\sqrt{73}$ units - Incorrect. The amplitude of the SHM component is 5. The maximum displacement from the origin is not simply related to $\sqrt{73}$ .
- D: The maximum displacement of the particle from the origin is 9 units - Correct. The maximum value of x is $4 + 5 = 9$ , and the minimum value is $4 - 5 = -1$ . So the maximum displacement from the origin is 9.

Why other options are wrong:
- Option A (The motion of the particle is S.H.M): The motion is SHM with a constant displacement, but the statement is technically incorrect as it doesn't account for the constant displacement.
- Option B (The amplitude of the S.H.M. of the particle is 5 units): This is correct.
- Option D (The maximum displacement of the particle from the origin is 9 units): This is correct.

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

Memory Aids:
- Remember the trigonometric identity for $\cos^2(x)$ .
- Remember how to combine sine and cosine terms into a single SHM.

Quick Tips:
- Use trigonometric identities to simplify complex expressions.
- Pay attention to constant displacements when analyzing SHM.

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Application +4 | -1

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are

{{K_1}}

and

{{K_2}}

. The time period of the suspended mass will be

View answer and explanation

Correct option: C

T = 2\pi \sqrt {\left( {\frac{{m\left( {{K_1} + {K_2}} \right)}}{{{K_1}{K_2}}}} \right)}

Option C (

T = 2\pi \sqrt {\left( {\frac{{m\left( {{{K_1} + {K_2}}}\right)}}{{{{K_1}{K_2}}}}} \right)}

) is correct because it uses the equivalent spring constant for springs in series to calculate the period.

Solution approach

Correct Answer + Core Reason: $T = 2\pi \sqrt {\left( {\frac{{m\left( {{{K_1} + {K_2}}}\right)}}{{{{K_1}{K_2}}}}} \right)}$ . The period is calculated using the effective spring constant for springs in series.

Step-by-Step Solution:

Step 1: When springs are connected in series, the effective spring constant $K_{eff}$ is given by $\frac{1}{K_{eff}} = \frac{1}{K_1} + \frac{1}{K_2}$ .

Step 2: Solve for $K_{eff}$ : $K_{eff} = \frac{K_1 K_2}{K_1 + K_2}$ .

Step 3: The time period of the suspended mass is given by $T = 2\pi \sqrt{\frac{m}{K_{eff}}}$ .

Step 4: Substitute $K_{eff}$ into the time period equation: $T = 2\pi \sqrt{\frac{m}{\frac{K_1 K_2}{K_1 + K_2}}} = 2\pi \sqrt{\frac{m(K_1 + K_2)}{K_1 K_2}}$ .

Why other options are wrong:
- Option A: Incorrect. This formula is for springs in parallel, not series.
- Option B: Incorrect. The factor of 2 is incorrect, and the spring constant is for parallel combination.
- Option D: Incorrect. The expression inside the square root is inverted.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Time period of a spring-mass system: $T = 2\pi \sqrt{\frac{m}{k}}$
- Effective spring constant for springs in series: $\frac{1}{K_{eff}} = \frac{1}{K_1} + \frac{1}{K_2}$

Memory Aids:
- Remember the formula for the effective spring constant in series and parallel.

Quick Tips:
- Always find the effective spring constant first when dealing with multiple springs.
- For springs in series, the effective spring constant is always less than the individual spring constants.

76 Medium

76 of 94Question permalink

Analyzing +4 | -1

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force.

F = {F_0}\sin \omega t

If the amplitude of the particle is maximum for ω=

{\omega _1}

and the energy of the particle is maximum for ω=

{\omega _2}

then

View answer and explanation

Correct option: C

{\omega _1} \ne {\omega _0}\,\,\mathrm{and}\,\,{\omega _2} = {\omega _0}

Option C (

{\omega _1} \ne {\omega _0}\,\mathrm{and}\,{\omega _2} = {\omega _0}

) is correct because the amplitude is maximum near the resonant frequency, but the energy is maximum exactly at the resonant frequency.

Solution approach

Correct Answer + Core Reason: ${\omega _1} \ne {\omega _0}\,\mathrm{and}\,{\omega _2} = {\omega _0}$ . Amplitude is maximum near resonance, energy is maximum at resonance.

Step-by-Step Solution:

Step 1: Understand the concept of resonance in forced oscillations. When a system is driven by an external force with a frequency close to its natural frequency, resonance occurs.

Step 2: The amplitude of the particle is maximum when the driving frequency is close to, but not exactly equal to, the natural frequency ( ${\omega _0}$ ). This is because the damping force affects the frequency at which the amplitude is maximized. Therefore, ${\omega _1} \ne {\omega _0}$ .

Step 3: The energy of the particle is maximum when the driving frequency is exactly equal to the natural frequency ( ${\omega _0}$ ). At resonance, the energy transfer from the driving force to the particle is most efficient. Therefore, ${\omega _2} = {\omega _0}$ .

Why other options are wrong:
- **Option A ( ${\omega _1} = {\omega _0}\,\mathrm{and}\,{\omega _2} \ne {\omega _0}$ ): This is incorrect because the amplitude is not exactly maximum at the natural frequency due to damping.
- Option B ( ${\omega _1} = {\omega _0}\,\mathrm{and}\,{\omega _2} = {\omega _0}$ ): This is incorrect because the amplitude is not exactly maximum at the natural frequency due to damping.
- Option D ( ${\omega _1} \ne {\omega _0}\,\mathrm{and}\,{\omega _2} \ne {\omega _0}$ ): This is incorrect because the energy is maximum at the natural frequency.

Key Concepts:
- Forced oscillations
- Resonance
- Natural frequency
- Damping

Memory Aids:
- Remember that resonance occurs when the driving frequency is close to the natural frequency.

Quick Tips:**
- In forced oscillations, the amplitude is maximum near the resonant frequency, while the energy is maximum at the resonant frequency.

77 Medium

77 of 94Question permalink

Understanding +4 | -1

The amplitude of vibration of a particle is given by

{a_m} = ({a_0})/(a{\omega ^2} - b\omega + c);

where

{a_0},a,b

and c are positive. The condition for a single resonant frequency is

View answer and explanation

Correct option: A

{b^2} = 4ac

Option A (

{b^2} = 4ac

) is correct because the condition for a single resonant frequency occurs when the denominator of the amplitude equation has a single minimum, which happens when the discriminant of the quadratic is zero.

Solution approach

Correct Answer + Core Reason: ${b^2} = 4ac$ . This is the condition for critical damping, leading to a single resonant frequency.

Step-by-Step Solution:

Step 1: Analyze the given amplitude equation: ${a_m} = \frac{{{a_0}}}{{a{\omega ^2} - b\omega + c}}$ .

Step 2: Recognize that resonance occurs when the denominator is minimized.

Step 3: The denominator is a quadratic expression in ω. For a single resonant frequency, the quadratic equation $a\omega^2 - b\omega + c = 0$ must have only one real root.

Step 4: The condition for a quadratic equation to have one real root (i.e., equal roots) is that its discriminant is zero: $D = b^2 - 4ac = 0$ .

Step 5: Therefore, the condition for a single resonant frequency is $b^2 = 4ac$ .

Why other options are wrong:
- Option B: ${b^2} > 4ac$ : This condition implies two distinct real roots, meaning there would be a range of resonant frequencies.
- Option C: ${b^2} = 5ac$ : This is an arbitrary condition and does not relate to the resonance condition.
- Option D: ${b^2} = 7ac$ : This is an arbitrary condition and does not relate to the resonance condition.

Key Concepts:
- Resonance
- Damped oscillations
- Quadratic equations
- Discriminant of a quadratic equation

Memory Aids:
- Remember that the discriminant ( $b^2 - 4ac$ ) determines the nature of the roots of a quadratic equation.

Quick Tips:
- For resonance problems, look for conditions that minimize the denominator of the amplitude equation.

78 Medium

78 of 94Question permalink

Understanding +4 | -1

A particle is executing S.H.M. Then the graph of acceleration as a function of displacement is

View answer and explanation

Correct option: A

A straight line

Option A (A straight line) is correct because acceleration in SHM is directly proportional to the negative of the displacement, resulting in a linear relationship.

Solution approach

Correct Answer + Core Reason: A straight line. Acceleration is directly proportional to displacement in SHM, $a = -\omega^2 x$ .

Step-by-Step Solution:

Step 1: Recall the equation for acceleration in SHM: $a = -\omega^2 x$ , where 'a' is acceleration, 'ω' is angular frequency, and 'x' is displacement.

Step 2: Recognize that this equation represents a linear relationship between acceleration (a) and displacement (x).

Step 3: The graph of a linear equation is a straight line.

Why other options are wrong:
- Option B: A circle: A circle would require a relationship like $a^2 + x^2 = constant$ .
- Option C: An ellipse: An ellipse would require a relationship like $\frac{a^2}{A^2} + \frac{x^2}{B^2} = 1$ , where A and B are different constants.
- Option D: A hyperbola: A hyperbola would require a relationship like $\frac{a^2}{A^2} - \frac{x^2}{B^2} = 1$ .

Key Concepts:
- Simple Harmonic Motion (SHM)
- Acceleration in SHM
- Linear relationship

Memory Aids:
- Remember that acceleration is proportional to the negative of displacement in SHM.

Quick Tips:
- Identify the mathematical relationship between the variables to determine the shape of the graph.

79 Medium

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Understanding +4 | -1

The potential energy of a particle (

{U_x}

) executing SHM is given by

View answer and explanation

Correct option: A

{U_x} = \frac{k}{2}{\left( {x - a} \right)^2}

Option A (

{U_x} = \frac{k}{2}{\left( {x - a} \right)^2}

) is correct because the potential energy in SHM is proportional to the square of the displacement from the equilibrium position.

Solution approach

Correct Answer + Core Reason: ${U_x} = \frac{k}{2}{\left( {x - a} \right)^2}$ . The potential energy of a particle in SHM is proportional to the square of its displacement from the equilibrium position.

Step-by-Step Solution:

Step 1: Recall the general form of potential energy in SHM.

Step 2: The potential energy $U(x)$ for a simple harmonic oscillator is given by $U(x) = \frac{1}{2}kx^2$ , where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. If the equilibrium position is at $x=a$ , then the displacement is $(x-a)$ .

Step 3: Substitute $(x-a)$ for $x$ in the potential energy equation: $U(x) = \frac{1}{2}k(x-a)^2$ .

Step 4: Rewrite the equation: ${U_x} = \frac{k}{2}{\left( {x - a} \right)^2}$

Why other options are wrong:
- Option B: ${U_x} = {k_1}x + {k_2}{x^2} + {k_3}{x^3}$ : This represents a more general potential energy function, not specifically SHM.
- Option C: ${U_x} = A{e^{ - bx}}$ : This is an exponential decay function, not related to SHM potential energy.
- Option D: ${U_x} = \mathrm{a constant}$ : This represents a constant potential energy, which would not result in oscillatory motion.

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

Memory Aids:
- Remember that potential energy in SHM is proportional to the square of the displacement.

Quick Tips:
- Recognize the quadratic form of potential energy in SHM.
- The equilibrium position is where the potential energy is minimum.

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Understanding +4 | -1

The bob of a simple pendulum of length L is released at time

t\mathrm{ } = \mathrm{ }0

from a position of small angular displacement. Its linear displacement at time t is given by

View answer and explanation

Correct option: D

\mathrm{X} = \mathrm{a cos}\sqrt {\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}

Option D (

\mathrm{X} = \mathrm{a cos}\sqrt {\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}

) is correct because the bob is released from a position of small angular displacement, implying a cosine function.

Solution approach

Correct Answer + Core Reason: $\mathrm{X} = \mathrm{a cos}\sqrt {\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}$ . Since the pendulum starts at maximum displacement, a cosine function is appropriate.

Step-by-Step Solution:

Step 1: Recognize that the linear displacement in SHM can be represented as $X = a \cos(\omega t)$ or $X = a \sin(\omega t)$ , where $a$ is the amplitude and $\omega$ is the angular frequency.

Step 2: Since the pendulum is released from a position of small angular displacement at $t=0$ , the displacement is maximum at $t=0$ . This corresponds to a cosine function.

Step 3: The angular frequency for a simple pendulum is given by $\omega = \sqrt{\frac{g}{L}}$ .

Step 4: Substitute the angular frequency into the cosine function: $X = a \cos(\sqrt{\frac{g}{L}} t)$ .

Why other options are wrong:
- Option A: $X = a\sin 2\pi \sqrt {\frac{L}{\mathrm{g}}} \times t$ . Incorrect, uses sine function and incorrect frequency.
- Option B: $X = a\cos 2\pi \sqrt {\frac{\mathrm{g}}{L}} \times t$ . Incorrect, uses incorrect frequency.
- Option C: $\mathrm{X} = \mathrm{a sin}\sqrt {\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}$ . Incorrect, uses sine function.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Angular frequency of a simple pendulum
- Cosine and sine functions in SHM

Memory Aids:
- Maximum displacement at t=0 implies cosine.

Quick Tips:
- Consider initial conditions to choose between sine and cosine.

81 Medium

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Application +4 | -1

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is

10{s^{ - 1}}

. At, t= 0 the displacement is 5m. What is the maximum acceleration? (The initial phase is

\frac{\pi }{4}

)

View answer and explanation

Correct option: B

500\sqrt 2 \,\,m/{s^2}

Option B (

500\sqrt 2 \,\,m/{s^2}

) is correct because it is the maximum acceleration calculated using the given ratio of maximum acceleration to maximum velocity and the initial displacement.

Solution approach

Correct Answer + Core Reason: $500\sqrt 2 \,\,m/{s^2}$ . The maximum acceleration is calculated using the given ratio and initial displacement.

Step-by-Step Solution:

Step 1: Given: $\frac{a_{max}}{v_{max}} = 10 s^{-1}$ , $x(0) = 5m$ , and initial phase $\phi = \frac{\pi}{4}$ .

Step 2: Recall the equations for displacement, velocity, and acceleration in SHM:\nx(t) = A \sin(ω t + φ)\nv(t) = Aω \cos(ω t + φ)\na(t) = -Aω^2 \sin(ω t + φ)

Step 3: Maximum velocity $v_{max} = A\omega$ and maximum acceleration $a_{max} = A\omega^2$ .\nSo, $\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega = 10 s^{-1}$ .

Step 4: At $t = 0$ , $x(0) = A \sin(\phi) = A \sin(\frac{\pi}{4}) = A \frac{1}{\sqrt{2}} = 5$ .\nTherefore, $A = 5\sqrt{2} m$ .

Step 5: Calculate the maximum acceleration: $a_{max} = A\omega^2 = (5\sqrt{2}) \times (10)^2 = 5\sqrt{2} \times 100 = 500\sqrt{2} \, m/s^2$ .

Why other options are wrong:
- Option A: $500\mathrm{ }\,m/{s^2}$ is incorrect as it does not match the correct calculation.
- Option C: $750\,m/{s^2}$ is incorrect as it does not match the correct calculation.
- Option D: $750\sqrt 2 \,\,m/{s^2}$ is incorrect as it does not match the correct calculation.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Displacement, velocity, and acceleration in SHM
- Amplitude
- Angular frequency
- Initial phase

Memory Aids:
- Remember the relationships between displacement, velocity, and acceleration in SHM.
- $v_{max} = A\omega$ and $a_{max} = A\omega^2$

Quick Tips:
- Use the given information to find the amplitude and angular frequency.
- Pay attention to the initial phase when calculating displacement, velocity, or acceleration at a specific time.

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Understanding +4 | -1

A particle is subjected simultaneously to two SHM’s one along the

x

-axis and the other along the

y

-axis. The two vibrations are in phase and have unequal amplitudes. The particle will execute

View answer and explanation

Correct option: A

Straight line motion

Option A (Straight line motion) is correct because when two SHMs are in phase, the resultant motion is a straight line with a slope equal to the ratio of their amplitudes.

Solution approach

Correct Answer + Core Reason: Straight line motion. When two SHMs are in phase, the resultant motion is a straight line.

Step-by-Step Solution:

Step 1: Let the two SHMs be represented as:
- $x = A_1 \sin(\omega t)$
- $y = A_2 \sin(\omega t)$ (since they are in phase)

Step 2: Since both SHMs have the same frequency and are in phase, we can write:
- $\frac{x}{A_1} = \sin(\omega t)$ and $\frac{y}{A_2} = \sin(\omega t)$

Step 3: Equate the two expressions:
- $\frac{x}{A_1} = \frac{y}{A_2}$

Step 4: Rearrange the equation to get:
- $y = \frac{A_2}{A_1} x$

Step 5: This equation represents a straight line passing through the origin with a slope of $\frac{A_2}{A_1}$ .

Why other options are wrong:
- Option B: Circular motion: This would require the two SHMs to have equal amplitudes and a phase difference of 90 degrees.
- Option C: Elliptic motion: This would require a phase difference between the two SHMs, but not necessarily 90 degrees, and unequal amplitudes.
- Option D: Parabolic motion: This would require a relationship like $y = kx^2$ , which is not the case here.

Key Concepts:
- Superposition of SHMs
- Phase difference
- Equation of a straight line

Memory Aids:
- Remember that in-phase SHMs result in straight-line motion.

Quick Tips:
- When dealing with superposition of SHMs, analyze the phase difference and amplitudes to determine the resulting motion.

83 Medium

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Analyzing +4 | -1

A particle is subjected to two simple harmonic motions along

{x}

and

{y}

directions according to

x=3 \sin 100 \pi t

and

y=4 \sin 100 \pi t

View answer and explanation

Correct option: B

Motion of particle will be on a straight line with slope 4/3

Option B (Motion of particle will be on a straight line with slope 4/3) is correct because both x and y are sinusoidal functions with the same frequency and no phase difference, indicating a straight-line motion.

Solution approach

Correct Answer + Core Reason: Motion of particle will be on a straight line with slope 4/3. Since both x and y have the same frequency and no phase difference, the motion is a straight line.

Step-by-Step Solution:

Step 1: Analyze the given equations: $x = 3 \sin(100\pi t)$ and $y = 4 \sin(100\pi t)$ .

Step 2: Notice that both x and y are sinusoidal functions with the same angular frequency ( $100\pi$ ) and no phase difference.

Step 3: Express $y$ in terms of $x$ : Since $\sin(100\pi t) = \frac{x}{3}$ , we can substitute this into the equation for $y$ : $y = 4 \left( \frac{x}{3} \right) = \frac{4}{3}x$ .

Step 4: Recognize that $y = \frac{4}{3}x$ is the equation of a straight line passing through the origin with a slope of $\frac{4}{3}$ .

Why other options are wrong:
- Option A (Motion of particle will be an ellipse traversing it in clockwise direction): This is incorrect because the motion is a straight line, not an ellipse. Elliptical motion occurs when there is a phase difference between the x and y motions.
- **Option C (Motion will be a simple harmonic motion along ${x}$ -axis with amplitude 5): This is incorrect because the motion is not just along the x-axis, but along a straight line in the x-y plane. Also, the amplitude is not simply 5.
- Option D (Phase difference between two motions is $\frac{\pi}{2}$ ): This is incorrect because there is no phase difference between the two motions. Both sine functions have the same argument.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Superposition of SHM
- Lissajous Figures

Memory Aids:
- If two SHMs have the same frequency and no phase difference, the resulting motion is a straight line.

Quick Tips:**
- When analyzing the superposition of two SHMs, check for the frequency and phase difference. If the frequencies are the same and there is no phase difference, the motion is a straight line.

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Analyzing +4 | -1

S-I: If the amplitude of the oscillation is doubled the maximum kinetic energy becomes four times.
S-II: If the amplitude of the oscillation is doubled the minimum potential energy also becomes four times.

View answer and explanation

Correct option: C

S-I is true

\&

S-II is false

Option C (S-I is true & S-II is false) is correct because maximum kinetic energy is proportional to the square of the amplitude, while minimum potential energy remains zero regardless of amplitude.

Solution approach

Correct Answer + Core Reason: S-I is true $\& S-II is false. Maximum kinetic energy is proportional to the square of the amplitude, while minimum potential energy is always zero.\n\n**Step-by-Step Solution:**\n\n**Step 1:** Statement I: If the amplitude of the oscillation is doubled the maximum kinetic energy becomes four times. This is true. The maximum kinetic energy is given by$ K_{max} = $\frac{1}{2}$ m ω^2 A² $. If$ A $is doubled,$ K_{max} $becomes four times.\n\n**Step 2:** Statement II: If the amplitude of the oscillation is doubled the minimum potential energy also becomes four times. This is false. The minimum potential energy in SHM is zero, which occurs at the equilibrium position, regardless of the amplitude.\n\n**Why other options are wrong:**\n- **Option A:** S-I$ \& S-II are true. This is incorrect because Statement II is false.
- Option B: S-I $\& S-II are false. This is incorrect because Statement I is true.\n- **Option D:** S-I is false$ \& S-II is true. This is incorrect because Statement I is true and Statement II is false.

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

Memory Aids:
- Remember that $K_{max} = \frac{1}{2} m \omega^2 A^2$ .
- Remember that the minimum potential energy in SHM is zero.

Quick Tips:
- Understand the relationship between amplitude and maximum kinetic energy.
- Know that the minimum potential energy in SHM is always zero.

85 Medium

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Understanding +4 | -1

The potential energy of a particle with displacement X is U(X). The motion is simple harmonic, when (K is a positive constant)

View answer and explanation

Correct option: A

U = \frac{{K{X^2}}}{2}

Option A (

U = \frac{{K{X^2}}}{2}

) is correct because the potential energy function for SHM is proportional to the square of the displacement.

Solution approach

Correct Answer + Core Reason: $U = \frac{{K{X^2}}}{2}$ . The potential energy function for simple harmonic motion is proportional to the square of the displacement.

Step-by-Step Solution:

Step 1: Recall the potential energy function for SHM.

Step 2: The potential energy $U$ for a simple harmonic oscillator is given by $U = \frac{1}{2}KX^2$ , where $K$ is a positive constant and $X$ is the displacement.

Step 3: Compare the given options with the standard form.

Step 4: Identify the option that matches the form $U = \frac{1}{2}KX^2$ .

Why other options are wrong:
- Option B: $U = K{X^2}$ : While proportional to $X^2$ , the factor of $\frac{1}{2}$ is missing from the standard form.
- Option C: $U = K$ : This represents constant potential energy, which doesn't result in SHM.
- Option D: $U = KX$ : This represents potential energy that is linearly proportional to displacement, which is not SHM.

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

Memory Aids:
- Remember that the potential energy in SHM is proportional to the square of the displacement.

Quick Tips:
- Recognize the quadratic form of potential energy in SHM.
- The restoring force is proportional to the negative gradient of the potential energy.

86 Medium

86 of 94Question permalink

Understanding +4 | -1

A body having natural frequency

\nu

' is executing forced oscillations under a driving force of frequency. The system will vibrate

View answer and explanation

Correct option: A

with frequency of driving force

\nu

Option A (with frequency of driving force

\nu

) is correct because in forced oscillations, the system vibrates at the frequency of the external driving force, not its natural frequency.

Solution approach

Correct Answer + Core Reason: with frequency of driving force $\nu$ . In forced oscillations, the system vibrates at the frequency of the driving force.

Step-by-Step Solution:

Step 1: Define forced oscillations: Oscillations where an external driving force causes the system to oscillate.

Step 2: In forced oscillations, the system will vibrate at the frequency of the driving force, regardless of its natural frequency.

Step 3: Therefore, the system will vibrate with frequency $\nu$ .

Why other options are wrong:
- Option B: with its natural frequency $v\mathrm{'}$ : This is incorrect because in forced oscillations, the system vibrates at the driving frequency, not its natural frequency.
- Option C: with mean frequency of the two $[\left( {v + v'} \right)/2$ : This is incorrect because the system vibrates at the driving frequency.
- Option D: None of the above: This is incorrect because the system vibrates at the driving frequency.

Key Concepts:
- Forced oscillations
- Natural frequency
- Driving frequency

Memory Aids:
- Forced oscillations vibrate at the driving frequency.

Quick Tips:
- Remember that in forced oscillations, the system vibrates at the driving frequency.

87 Medium

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Understanding +4 | -1

The restoring force of SHM is maximum when particle

View answer and explanation

Correct option: A

Displacement is maximum

Option A (Displacement is maximum) is correct because the restoring force is directly proportional to the displacement from the equilibrium position.

Solution approach

Correct Answer + Core Reason: Displacement is maximum. The restoring force is directly proportional to the displacement in SHM.

Step-by-Step Solution:

Step 1: Recall the definition of the restoring force in SHM: $F = -kx$ , where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.

Step 2: The magnitude of the restoring force is $|F| = k|x|$ .

Step 3: The restoring force is maximum when the displacement $|x|$ is maximum.

Why other options are wrong:
- Option B: Is half way between the mean and extreme position. Incorrect, the force is less than maximum here.
- Option C: Crosses mean position. Incorrect, the displacement and force are zero at the mean position.
- Option D: Is at rest. Incorrect, the force depends on displacement, not velocity.

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

Memory Aids:
- Force is proportional to displacement.

Quick Tips:
- Remember the relationship between restoring force and displacement.

88 Medium

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Understanding +4 | -1

The time period of a simple pendulum, when it is made to oscillate on the surface of moon

View answer and explanation

Correct option: A

Increases

Option A (Increases) is correct because the acceleration due to gravity on the moon is less than that on Earth, leading to a larger time period.

Solution approach

Correct Answer + Core Reason: Increases. Since the acceleration due to gravity on the moon is less than on Earth, the time period increases.

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: The acceleration due to gravity on the moon ( $g_m$ ) is approximately $\frac{1}{6}$ of the acceleration due to gravity on Earth ( $g_e$ ).

Step 3: Therefore, $g_m < g_e$ .

Step 4: Since $T \propto \frac{1}{\sqrt{g}}$ , a smaller value of $g$ results in a larger value of $T$ .

Step 5: Thus, the time period of the pendulum increases on the moon.

Why other options are wrong:
- Option B: Decreases: This is incorrect because the time period increases due to the lower gravity on the moon.
- Option C: Remains unchanged: This is incorrect because the time period depends on the acceleration due to gravity.
- Option D: Becomes infinite: This is incorrect because the time period increases but remains finite.

Key Concepts:
- Simple pendulum
- Time period of a simple pendulum
- Acceleration due to gravity on the moon

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

Quick Tips:
- Remember that the time period is inversely proportional to the square root of the acceleration due to gravity.

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Remembering +4 | -1

If a body oscillates at the angular frequency

{\omega _d}

of the driving force, then the oscillations are called

View answer and explanation

Correct option: C

Forced oscillations

The correct answer is C (Forced oscillations) because forced oscillations occur when a body oscillates at the frequency of the external driving force.

Solution approach

Correct Answer + Core Reason: Forced oscillations. When a body oscillates at the frequency of the driving force, it's called forced oscillation.

Step-by-Step Solution:

Step 1: Define free oscillations: Free oscillations occur when a system oscillates at its natural frequency without any external force.

Step 2: Define coupled oscillations: Coupled oscillations involve two or more oscillating systems that exchange energy.

Step 3: Define forced oscillations: Forced oscillations occur when an external periodic force is applied to a system, causing it to oscillate at the frequency of the driving force.

Step 4: Define maintained oscillations: Maintained oscillations are oscillations where energy lost due to damping is replenished to maintain a constant amplitude.

Step 5: In this case, the body oscillates at the angular frequency ${\omega _d}$ of the driving force. This is the definition of forced oscillations.

Why other options are wrong:
- Option A: Free oscillations occur at the natural frequency of the system, not the driving frequency.
- Option B: Coupled oscillations involve multiple oscillating systems, not just one driven by an external force.
- Option D: Maintained oscillations involve replenishing energy to counteract damping, but the frequency is still typically the natural frequency or a related frequency, not necessarily the driving frequency.

Key Concepts:
- Free Oscillations
- Forced Oscillations
- Coupled Oscillations
- Maintained Oscillations

Memory Aids:
- Forced oscillations: Think of "forced" to remember that the oscillation is driven by an external force.

Quick Tips:
- Understand the definitions of different types of oscillations to quickly identify the correct answer.

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Understanding +4 | -1

What is the effect on the time period of a simple pendulum if the mass of the bob is doubled

View answer and explanation

Correct option: D

No effect

Option D (No effect) is correct because the time period of a simple pendulum is independent of the mass of the bob.

Solution approach

Correct Answer + Core Reason: No effect. The time period of a simple pendulum is independent of the mass of the bob.

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: Observe that the formula does not include the mass of the bob.

Step 3: Therefore, changing the mass of the bob will have no effect on the time period of the pendulum.

Why other options are wrong:
- Option A: Halved: This is incorrect because the time period is independent of the mass.
- Option B: Doubled: This is incorrect because the time period is independent of the mass.
- Option C: Becomes eight times: This is incorrect because the time period is independent of the mass.

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

Memory Aids:
- The time period of a simple pendulum depends only on its length and the acceleration due to gravity.

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

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