Physics Questions and Answers – Oscillations – Energy in Simple Harmonic Motion


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

Answer: a
Explanation: The kinetic energy of a particle in SHM is given by:
0.5mA2w2cos2(wt + a), where a is the phase constant.
cos2(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

Answer: b
Explanation: Let x = Asin(wt).
∴ Kinetic energy = 1/2 mA2w2cos2(wt)
potential energy = 1/2kA2sin2(wt).
According to given condition:
1/2mA2w2cos2(wt) = 1/2A2w2sin2(wt)
using, k = mw2, we get:
tan2(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

Answer: a
Explanation: w = 2π/T = π.
Maximum kinetic energy = 1/2mA2w2, 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 = 1/2mA2w2 – 5
= (0.5*2*1*9.87) – 5
= 4.87J.
Note: Join free Sanfoundry classes at Telegram or Youtube

4. A particle is undergoing SHM. If its potential energy is given by: U = kx2. 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

Answer: a
Explanation: Let x = Asin(wt).
U = kA2sin2(wt)
K = 1/2mA2w2cos2(wt).
Given: U = 1/3K.
kA2sin2(wt) = (1/3)*(1/2)mA2w2cos2(wt),
and k = mw2.
∴ tan2(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

Answer: a
Explanation: Total energy = 1/2mw2A2.
So, if mass is doubled keeping the amplitude same,
total energy will get doubled.
Take Physics - Class 11 Mock Tests - Chapterwise!
Start the Test Now: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

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

Answer: a
Explanation: 1/2ka2 = 8 & 1/2kb2 = 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

Answer: a
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 = x2-8x+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

Answer: a
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 – x2)04 = v2/2.
∴ v = 4√2 m/s.

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


To practice all areas of Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.