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NEET • PHYSICS • OSCILLATIONS

Oscillations practice · Medium

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91 Medium

91 of 94Question permalink

Understanding +4 | -1

A particle of mass m is hanging vertically by an ideal spring of force constant K. If the mas is made to oscillate vertically, its total energy is

View answer and explanation

Correct option: D

Same at all position

Option D (Same at all position) is correct because the total energy in SHM is conserved and remains constant throughout the oscillation.

Solution approach

Correct Answer + Core Reason: Same at all position. The total energy in SHM is conserved.

Step-by-Step Solution:

Step 1: In a simple harmonic motion, the total energy is the sum of the potential energy and the kinetic energy.

Step 2: The potential energy is maximum at the extreme positions and minimum at the mean position.

Step 3: The kinetic energy is maximum at the mean position and minimum at the extreme positions.

Step 4: The total energy remains constant throughout the motion, as energy is continuously exchanged between potential and kinetic forms.

Step 5: Therefore, the total energy is the same at all positions.

Why other options are wrong:
- Option A: Maximum at extreme position is incorrect because the total energy is constant, not maximum at the extreme position.
- Option B: Maximum at mean position is incorrect because the total energy is constant, not maximum at the mean position.
- Option C: Minimum at mean position is incorrect because the total energy is constant, not minimum at the mean position.

Key Concepts:
- Simple harmonic motion (SHM)
- Conservation of energy

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

Quick Tips:
- The total energy in SHM is conserved and is the same at all positions.

92 Medium

92 of 94Question permalink

Analyzing +4 | -1

The displacement of a particle along the

x

axis is given by

x = a{\sin ^2}\omega t

The motion of the particle corresponds to

View answer and explanation

Correct option: D

Non simple harmonic motion

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

8\pi

m/s.

Solution approach

Correct Answer + Core Reason: Magnitude of the velocity of particle is 8 meter/second. The magnitude of the velocity is actually $8\pi$ m/s.

Step-by-Step Solution:

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

Step 2: Velocity $\overrightarrow{v} = \frac{d\overrightarrow{R}}{dt} = 8\pi \cos(2\pi t)\hat{I} - 8\pi \sin(2\pi t)\hat{J}$ .

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

Step 4: Acceleration $\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = -16\pi^2 \sin(2\pi t)\hat{I} - 16\pi^2 \cos(2\pi t)\hat{J} = -4\pi^2(4\sin(2\pi t)\hat{I} + 4\cos(2\pi t)\hat{J}) = -4\pi^2 \overrightarrow{R}$ .

Step 5: Magnitude of acceleration $|\overrightarrow{a}| = \sqrt{(-16\pi^2 \sin(2\pi t))^2 + (-16\pi^2 \cos(2\pi t))^2} = 16\pi^2$ . Also, $|\overrightarrow{a}| = \frac{v^2}{R} = \frac{(8\pi)^2}{4} = \frac{64\pi^2}{4} = 16\pi^2$ .

Step 6: The path of the particle is a circle of radius 4 meters because $|\overrightarrow{R}| = \sqrt{(4\sin(2\pi t))^2 + (4\cos(2\pi t))^2} = \sqrt{16} = 4$ .

Step 7: Final Answer: Magnitude of the velocity of particle is 8 meter/second

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

Key Concepts:
- Circular Motion
- Velocity and Acceleration in Circular Motion
- Position Vector

Memory Aids:
- Remember that the derivative of sine is cosine, and the derivative of cosine is negative sine.

Quick Tips:
- Differentiate the position vector to find the velocity vector.
- Differentiate the velocity vector to find the acceleration vector.
- Calculate the magnitudes of the vectors using the Pythagorean theorem.

93 Medium

93 of 94Question permalink

Analyzing +4 | -1

The displacement y of a particle executing periodic motion is given by

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

. This expression may be considered to be a result of the superposition of…….. independent harmonic motions

View answer and explanation

Correct option: B

Three

Option B (Three) is correct because the given equation can be expanded using trigonometric identities into three independent harmonic motions.

Solution approach

Correct Answer + Core Reason: Three. The given equation can be expanded into three independent harmonic motions using trigonometric identities.

Step-by-Step Solution:

Step 1: Analyze the given equation: $y = 4{\cos ^2}\left( {t/2} \right)\sin \left( {1000t} \right)$ .

Step 2: Use the trigonometric identity: ${\cos ^2}\theta = \frac{1 + \cos 2\theta}{2}$ .

Step 3: Substitute this identity into the given equation: $y = 4\left( {\frac{1 + \cos t}{2}} \right)\sin \left( {1000t} \right) = 2\left( {1 + \cos t} \right)\sin \left( {1000t} \right)$ .

Step 4: Expand the equation: $y = 2\sin \left( {1000t} \right) + 2\cos t\sin \left( {1000t} \right)$ .

Step 5: Use the trigonometric identity: $2\cos A\sin B = \sin (A + B) - \sin (A - B)$ .

Step 6: Apply this identity to the second term: $2\cos t\sin \left( {1000t} \right) = \sin (t + 1000t) - \sin (t - 1000t) = \sin (1001t) - \sin (-999t) = \sin (1001t) + \sin (999t)$ .

Step 7: Substitute this back into the equation: $y = 2\sin \left( {1000t} \right) + \sin (1001t) + \sin (999t)$ .

Step 8: The equation now represents the superposition of three independent harmonic motions with frequencies 1000, 1001, and 999.

Why other options are wrong:
- Option A (Two): This is incorrect because the equation expands into three terms.
- Option C (Four): This is incorrect because the equation expands into three terms.
- Option D (Five): This is incorrect because the equation expands into three terms.

Key Concepts:
- Superposition of harmonic motions
- Trigonometric identities

Memory Aids:
- Remember the trigonometric identities: ${\cos ^2}\theta = \frac{1 + \cos 2\theta}{2}$ and $2\cos A\sin B = \sin (A + B) - \sin (A - B)$ .

Quick Tips:
- When dealing with trigonometric functions in SHM, use trigonometric identities to simplify the equation.

94 Medium

94 of 94Question permalink

Application +4 | -1

A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system if a mass 4 m is suspended from the same spring

View answer and explanation

Correct option: C

n/2

Option C (

n/2

) is correct because the frequency is inversely proportional to the square root of the mass.

Solution approach

Correct Answer + Core Reason: $n/2$ . The frequency is inversely proportional to the square root of the mass.

Step-by-Step Solution:

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

Step 2: When the mass is changed to $4m$ , the new frequency $n'$ is given by $n' = \frac{1}{2\pi} \sqrt{\frac{k}{4m}}$ .

Step 3: Divide the new frequency by the original frequency: $\frac{n'}{n} = \frac{\frac{1}{2\pi} \sqrt{\frac{k}{4m}}}{\frac{1}{2\pi} \sqrt{\frac{k}{m}}} = \sqrt{\frac{k}{4m} \times \frac{m}{k}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$ .

Step 4: Therefore, $n' = \frac{n}{2}$ .

Why other options are wrong:
- Option A: Incorrect. The frequency is halved, not divided by 4.
- Option B: Incorrect. The frequency decreases, not increases by a factor of 4.
- Option D: Incorrect. The frequency is halved, not doubled.

Key Concepts:
- Simple Harmonic Motion (SHM)
- Frequency of oscillation of a spring-mass system: $n = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Memory Aids:
- Remember that frequency is inversely proportional to the square root of the mass.

Quick Tips:
- When the mass increases, the frequency decreases, and vice versa.

Revision notes and key takeaways

Review OSCILLATIONS in your textbook or class notes, then use this set to check your understanding.

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