This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Oscillations – Energy in Simple Harmonic Motion”.

1. What is the time period of kinetic energy of a particle in SHM,if the time period of SHM is 4s?

a) 2s

b) 1s

c) 4s

d) 0s

View Answer

Explanation: The kinetic energy of a particle in SHM is given by:

0.5mA

^{2}w

^{2}cos

^{2}(wt + a), where a is the phase constant.

cos

^{2}(wt + a) = (1 + cos2(wt+a))/2.

Therefore the time period of oscillation of kinetic energy is 4/2 = 2s.

2. Consider a particle undergoing an SHM of amplitude A & angular frequency w. What is the magnitude of displacement from the mean position when kinetic energy is equal to the magnitude of potential energy?

a) A/2

b) A/√2

c) A/3

d) A/4

View Answer

Explanation: Let x = Asin(wt).

∴ Kinetic energy = 1/2 mA

^{2}w

^{2}cos

^{2}(wt)

potential energy = 1/2kA

^{2}sin

^{2}(wt).

According to given condition:

1/2mA

^{2}w

^{2}cos

^{2}(wt) = 1/2A

^{2}w

^{2}sin

^{2}(wt)

using, k = mw

^{2}, we get:

tan

^{2}(wt) = 1.

∴ wt = π/4.

∴ x = Asin(π/4) = A/√2.

3. The graph given below represents energy as a function of position. What is the energy of the particle in SHM at x = +1? Given mass of the particle is 2kg, time period =2s, A=1m.

a) 4.87J

b) -5J

c) 9.87J

d) 5J

View Answer

Explanation: w = 2π/T = π.

Maximum kinetic energy = 1/2mA

^{2}w

^{2}, which will be at x=0.

Total energy will be constant,

therefore energy at x=0 is the same as the energy at x = +1.

Thus, energy = 1/2mA

^{2}w

^{2}– 5

= (0.5*2*1*9.87) – 5

= 4.87J.

4. A particle is undergoing SHM. If its potential energy is given by: U = kx^{2}. What will be the value of x for potential energy to be 1/3rd of kinetic energy? Assume k is force constant.

a) A/√7

b) A/√6

c) A/√3

d) A/√2

View Answer

Explanation: Let x = Asin(wt).

U = kA

^{2}sin

^{2}(wt)

K = 1/2mA

^{2}w

^{2}cos

^{2}(wt).

Given: U = 1/3K.

kA

^{2}sin

^{2}(wt) = (1/3)*(1/2)mA

^{2}w

^{2}cos

^{2}(wt),

and k = mw

^{2}.

∴ tan

^{2}(wt) = 1/6. Or tan(wt) = 1/√6.

∴ sin(wt) = 1/√7.

x = Asin(wt) = A/√7.

5. A particle is performing a SHM. If its mass is doubled keeping the amplitude and force constant the same, total energy will become how many times the initial value?

a) 2

b) 1/2

c) 4

d) 1

View Answer

Explanation: Total energy = 1/2mw

^{2}A

^{2}.

So, if mass is doubled keeping the amplitude same,

total energy will get doubled.

6. A particle is in SHM. When displacement is ‘a‘ potential energy is 8J. When displacement is ‘a‘ potential energy is 18J. What will be the potential energy when displacement is ‘a+b’?

a) 50J

b) 26J

c) 10J

d) 30J

View Answer

Explanation: 1/2ka

^{2}= 8 & 1/2kb

^{2}= 18.

We have to find value of 1/2k(a + b)

^{2}.

∴ 1/2k(a + b)

^{2}= 8 + 18 + kab

= 26+(k*4*6/k)

= 26 + 24 = 50J.

7. If the total energy of a particle in SHM is 200J, the value of kinetic energy at an instant can be 220J. True or False?

a) True

b) False

View Answer

Explanation: The total energy of the particle is the sum of kinetic & potential energy. Potential energy is frame dependent, so it can be negative in some cases. Thus, the value of kinetic energy can exceed total energy.

8. If the potential energy of the particle is given by: U = x^{2}-8x+16. It starts from rest from x=0. What will be its maximum speed during SHM? Assume the mass of the particle to be 1kg.

a) 4√2m/s

b) 8m/s

c) 2√2m/s

d) 2√6m/s

View Answer

Explanation: F = -dU/dx = 8 – 2x.

Particle starts from x=0, so force will increase its speed till force becomes zero at x=4.

Let acceleration be ‘a’.

a = dv/dt = v dv/dx.

And a = F/m = F/1 = 8-2x.

adx = vdv. Now, integrate both sides. The limits of dx will be 0 to 4 and the limits of dv will be 0 to v.

(8x – x

^{2})

_{0}

^{4}= v

^{2}/2.

∴ v = 4√2 m/s.

**Sanfoundry Global Education & Learning Series – Physics – Class 11**.

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