# Class 11 Physics MCQ – Oscillations – Simple Harmonic Motion and Uniform Circular Motion

This set of Class 11 Physics Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Oscillations – Simple Harmonic Motion and Uniform Circular Motion”.

1. In the given phasor diagram what should be the minimum velocity of particle 2 at mean position so that it catches particle 1 before it reaches the extreme position? The projection of motion of particle 1 on x-axis is governed by the equation: x = 4sin(2t).

a) 8m/s
b) 16m/s
c) 4m/s
d) 12m/s
View Answer

Answer: b
Explanation: 2 has to catch 1 before it covers a quarter of the circle. In this time, 2 will have to cover half the circle. Thus, it should have twice the velocity of particle 1 at each corresponding point. Thus, minimum velocity of particle 2 at mean position is twice that of 1 at mean position
= 2*( 4*2)
= 16m/s.

2. A ball tied to the end of a string of length 10cm is rotated with constant speed of 2m/s in a circle about the origin. Find the equation of projection of the ball on the x-axis. At t=0, the ball makes an angle of 45° with the x-axis & is travelling in anticlockwise direction.
a) 0.1sin(20t + 3π/4)
b) 0.1cos(20t + π/4)
c) 10sin(20t + π/4)
d) 10cos(20t – 3π/4)
View Answer

Answer: a
Explanation: Let the equation of motion be: x = A sin(wt + a). The amplitude is 0.1m, w = 2/0.1 = 20s-1. a will be either π/4 or 3π/4 since at t=0, x = A/√2. The ball is travelling in anticlockwise direction so its velocity of projection at t=0 should be towards the origin and therefore negative. v = Aw cos(wt + a).
At t=0, v is negative, so a will be 3π/4. Therefore eqn is: x = 0.1sin(20t + 3π/4)

3. A ball is hanging from a thread. It is given some speed such that the ball makes a complete circle under the influence of gravity. The projection of the ball on the x-axis will be an SHM. True or False?

a) True
b) False
View Answer

Answer: b
Explanation: For the projection of a circular motion to be a SHM, the speed of a particle in circular motion should be constant, but under the influence of gravity, the speed will be variable. Thus, the motion of projection will be oscillatory but not simple harmonic.
advertisement
advertisement

4. If a body in uniform circular motion covers the given angle in 0.5s, what is the time period of projection, on the x-axis, undergoing SHM? Assume it moves anti-clockwise.

a) 4s
b) 1s
c) 2s
d) 3s
View Answer

Answer: a
Explanation: The given motion represents the body moving from A/√2 to 0.
The time taken from 0 to A/√2 will also be the same.
wt = angle covered by circular motion, where w is angular frequency = 2π/T.
w*0.5 = π/4. w = π/2.
T = 2π/w = 4s.

5. The angular speed of a body in uniform circular motion is the same as the angular frequency of SHM of its projection. True or False?
a) True
b) False
View Answer

Answer: a
Explanation: Let the angular speed of the body be w. Let it start from rest from the x-axis.
At any time the angle it has covered is wt.
The projection on x-axis is then given by: x = Asin(wt).
Thus, angular speed is the same as angular frequency of SHM.

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

To practice all chapters and topics of class 11 Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement

advertisement
advertisement
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, 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 & discussions at Telegram SanfoundryClasses.