Simple Harmonic Motion And Uniform Circular Motion MCQ - Practice Questions with Answers

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean position is separated by distance $X_0\left(X_0>A\right)$. If the maximum separation between them is $\left(X_0+A\right)$, the phase difference between their motion is

A simple pendulum performs simple harmonic motion about $x=0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $x=\frac{A}{2}$ will be

The circle's radius, the period of revolution, the initial position and sense of revolution are indicated in Fig. 

Y-projection of the radius vector of rotating particle P is:

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  $\frac{-A}{2}$    and they are moving towards each other.  If they cross each other at time t, then t is :

The radius of the circle, the Period of revolution, the initial Position, and the sense of revolution are indicated in Figure Y. The Projection of the radius vector of a rotating particle P is: