Physics Multiple Choice Questions 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
Answer: b
Clarification: 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)
Answer: a
Clarification: 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
Answer: b
Clarification: 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.
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
Answer: a
Clarification: 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
Answer: a
Clarification: 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.