Energy in SHM

Practice Questions:

Question 1: difficult

For a particle executing simple harmonic motion, the displacement x is given by x = A cosωt. Identify the graph which represents the variation of potential energy (PE) as function of time t and displacement x

1. I, III

2. II, IV

3. II, III

4. I, IV

View Answer

\[ P.E = \frac{1}{2}Kx^{2} = \frac{1}{2}KA^{2}cos^{2}\left( \omega t \right) \]

Graph I represents graph of cos²ωt.

\[ P.E = \frac{1}{2}Kx^{2} \]

Graph III represents a parabolic function

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Question 2: moderate

A body is executing simple harmonic motion. At a displacement x, its potential energy is E1 and at a displacement y, its potential energy is E2. The potential energy E at a displacement (x + y) is :

1. E1 + E2

2. √E1² + E2²

3. E1 + E2 + 2√E1E2

4. √E1E2

View Answer

\[ E_{1}= \frac{1}{2}Kx^{2} \]

\[ E_{2}= \frac{1}{2}Ky^{2} \]

\[ E= \frac{1}{2}K(x+y)^{2}= \frac{1}{2}Kx^{2} + \frac{1}{2}Ky^{2} + Kxy \]

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Question 3: moderate

A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 Nm–¹. The block is pulled to a distance of x = 10 cm from its equilibrium position at x = 0 cm on a frictionless surface from rest at t = 0. The kinetic energy of the block when it is 5 cm away from the mean position is

1. 0.12 J

2. 0.15 J

3. 0.19 J

4. 0.21 J

View Answer

\[ K.E= \frac{1}{2}K\left( A^{2}-x^{2} \right)\]

\[ K.E= \frac{1}{2}\times 50\left( 10^{2}-5^{2} \right)/10^{4}= 0.19 J \]

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Question 4: moderate

The total mechanical energy of a spring-mass system in simple harmonic motion is E=1/2mω²A². Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will :

1. become 2E

2. become √2E

3. become E/2

4. remain E

View Answer

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Question 5: moderate

The total mechanical energy of a particle executing simple harmonic motion is E. When the displacement is half the amplitude its kinetic energy will be :

1. 3E/4

2. E

3. E/2

4. E/4

View Answer

\[ K.E= \frac{1}{2}K\left(A^{2}-\left( \frac{A}{2}\right)^{2} \right)= \frac{3}{8} K A^{2} \]

K.E= 3E/4

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Question 6: moderate

A particle is executing S.H.M., If its P.E. & K.E. is equal then the ratio of displacement & amplitude will be :

1. 1/√2

2. √2

3. 1/2

4. 3/2

View Answer

\[ K.E= \frac{1}{2}K\left(A^{2}-x^{2} \right) \]

\[ P.E= \frac{1}{2}Kx^{2} \]

\[ \frac{1}{2}K\left(A^{2}-x^{2} \right)= \frac{1}{2}Kx^{2} \]

\[ x= A/\sqrt{2} \]

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Question 7: moderate

A particle is executing linear simple harmonic motion of amplitude A. What fraction of the total energy is kinetic when the displacement is half the amplitude

1. 1/4

2. 1/2√2

3. 1/2

4. 3/4

View Answer

\[ K.E= \frac{1}{2}K\left(A^{2}-x^{2} \right) \]

\[ K.E= \frac{1}{2}K\left(A^{2}-\left( \frac{A}{2}\right)^{2} \right)= \frac{3}{8} K A^{2} \]

K.E/T.E =3/4

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Question 8: moderate

A particle of mass 0.1 kg executes SHM under a force F = (–10x) Newton. Speed of particle at mean position is 6 m/s. Then amplitude of oscillations is :

1. 0.6 m

2. 0.2 m

3. 0.4 m

4. 0.1 m

View Answer

Spring Constant K = m ω² ⇒ 10 = 0.1 ω² ⇒ ω²= 100 ⇒ ω = 10 rad/sec

Speed is maximum at mean position

Vmax= Aω

6= A × 10

A = 0.6 m

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Question 9: moderate

The potential energy of a particle executing simple harmonic motion at a distance x from the equilibrium position is proportional to :

1. √x

2. x

3. x²

4. x³

View Answer

\[ U = \frac{1}{2}kx^{2} \]

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Question 10: easy

What should be the displacement of a simple pendulum whose amplitude is A, at which potential energy is 1/4 th of the total energy ?

1. A/√2

2. A/2

3. A/4

4. A/2√2

View Answer

\[ \frac{1}{2} k x^{2}=\frac{1}{4}\left( \frac{1}{2}kA^{2} \right) \]

\[ \frac{1}{2}k x^{2}=\frac{1}{4}\left(kA^{2} \right) x= \frac{A}{2} \]

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