Question

A particle is executing a vertical SHM about the highest point of the projectile when the particle is at the mean position; the projectile is fired from the ground with velocity u at an angle θ with the horizontal. The projectile hits the oscillating particle. Then, the possible time period of the particle is:
$\begin{align}
  & A.\dfrac{u\sin \theta }{g} \\
 & B.\dfrac{2u\sin \theta }{g} \\
 & C.\dfrac{2u\sin \theta }{3g} \\
 & D.\text{All the above} \\
\end{align}$

Answer 1

Hint: Projectile motion is a motion in which an object is thrown with some initial velocity. Find the time for a maximum height of the projectile by applying the concept of simple harmonic motion, then evaluate the expression for the final answer.
Formula used: $t=\dfrac{u\sin \theta }{g}$

Complete answer:
We know that simple harmonic motion or SHM is defined as the motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards the mean position. So the direction of the acceleration is opposite to the direction of the motion. The time to reach the highest point is given by the following expression:
 $t=\dfrac{u\sin \theta }{g}......\left( 1 \right)$
Where ‘u’ is the velocity of the projectile, ‘g’ is the acceleration due to gravity and θ is the angle made by the projectile with the horizontal. It is given that particle is executing SHM, it will pass through its mean position at times,
$t=0,\dfrac{T}{2},T,\dfrac{3T}{2},.......$
In general
$t=\dfrac{nT}{2}......\left( 2 \right)$
Where, n=0, 1, 2……
By evaluating (1) and (2) we get,
$\begin{align}
  & \dfrac{u\sin \theta }{g}=\dfrac{nT}{2} \\
 & T=\dfrac{2u\sin \theta }{ng} \\
\end{align}$
Where, n= 1, 2, 3…..
So, as we know T can become
$T=\dfrac{2u\sin \theta }{g},\dfrac{u\sin \theta }{g},\dfrac{2u\sin \theta }{3g},.....$

Therefore, the correct option for the given question is (d).

Note:
SHM is a special case of oscillatory motion. In which an object will keep on moving between two extreme positions about a fixed position is called mean position. At the mean position, the net force on the particle is zero. Therefore, it is a stable equilibrium position. The particle will pass through the mean position having SHM only at the time.
$t=0,\dfrac{T}{2},T,\dfrac{3T}{2},2T,\text{up to so on}.$