MEDIUM
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Explain the relationship between uniform circular motion and SHM.

Important Questions on Oscillations

MEDIUM
Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement -A2 and they are moving towards each other. If they cross each other at time t, then t is:
EASY

For particle P revolving round the centre O with radius of circular path r and regular velocity ω, as shown in below figure, the projection of OP on the x-axis at time t is

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EASY
If we tie a stone to the end of a string and move it with a constant angular speed in a horizontal plane about fixed point, the stone would perform a : 
MEDIUM
A stone is swinging in a horizontal circle 0.8 m in diameter, at 30 rev min-1. A distant light causes a shadow of the stone to be formed on a nearby wall. What is the amplitude of the motion of the shadow? What is the frequency?
EASY
A particle is in a uniform circular motion around origin. What is the angle made by the position vector of the particle with the X-axis, when its displacement along the X-axis becomes half the maximum displacement?
HARD
The x-projection for a certain particle in circular motion under SHM with period of 2 s, amplitude of oscillation is 2 m and initial phase of π2 is
EASY
A stone is swinging in a horizontal circle 1 m in diameter at 60 rev min-1. A distant light causes a shadow of the stone to be formed on a nearly vertical wall. The amplitude and period of motion for the shadow of the stone are,
MEDIUM
A particle executes SHM with time period T and amplitude A. The maximum possible average velocity in time T/4 is
EASY
A particle executes simple harmonic motion between x=-A and x=+A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then
EASY
A particle is revolving in a circular path with gradually increasing speed. Its motion is
MEDIUM
If the particle is moving in circular motion under SHM, then its x-projection is depending upon
EASY
The value of phase at maximum displacement from the mean position of a particle in S.H.M. is
EASY
Suppose a particle P is moving uniformly on a circle of radius A with the angular speed ω . The sense of rotation is anticlockwise. If the  t=0 , it  makes an angle of ϕ with the positive direction of the x-axis. In time t, it will cover a further angle ωt.What is the projection of position vector on the X-axis at time t.
EASY
A particle executes simple harmonic motion between x=-A and x=+A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then
MEDIUM
Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement -A2 and they are moving towards each other. If they cross each other at time t, then t is:
MEDIUM

Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is : 

(Given, time period of revolution T=30 s)

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HARD

A particle is moving in a uniform circular motion on a horizontal surface. Particle's position and velocity at time t=0 are shown in the figure in the coordinate system. Which of the indicated variable on the vertical axis is/are correctly matched by the graph(s) shown alongside for particle's motion?

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EASY

A particle is moving in a uniform circular motion on a horizontal surface. Particle position and velocity at time t=0 are shown in the figure in the coordinate system. Which of the indicated variable on the vertical axis is incorrectly matched by the graph shown alongside for particle's motion-
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MEDIUM
A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 sec. The minimum time taken by the particle to move from x=+2 cm to x=+4 cm and back again is given by: