Question

A ball resting on a tray attached to a spring is made to oscillate vertically, such that the fixed end of the spring is attached to the ground. If the spring is compressed and let go, the ball will execute SHM until

A. the ball is in contact with the spring
B. the ball occasionally loses contact with the spring
C. the ball loses contact with the spring beyond the mean position
D. the ball loses contact with the spring beyond extreme position

Answer 1

At first we need to understand and analyze the problem properly. Then we must draw a figure that will help us to find the actual condition of the question. Now if a body is going through SHM then anything that is in contact with the body will also be showing SHM. When the ball is kept on the tray it will compress the string and the string will also try to give an opposite restoring force.

Complete step-by-step answer :
We have to understand the problem at first, in the question it is written that a tray is in one side of the spring and another side is attached to a fixed point that is to the ground, Now a ball is placed then the tray, and it is compressed and released then the ball will execute SHM till that time.

In the above diagram we see a 2-D representation of the question that is given. Here we see a ball on the tray which is fixed with a spring and the spring is fixed at a point on the other end.
Now, according to the question, when we place the ball on the spring, the ball has a mass of its own due to which the spring gets compressed to a amount that we can consider as ${{y}_{0}}$, we know that gravitational force is also acting on it and hence the force applied by the spring will be, $mg=k{{y}_{0}}$
Now, as the spring is compressed due to the weight of the ball then a restoring force will act on the spring that will be in the upward direction.
Therefore the total force that will be acting when the spring is fully compressed will be $mg-k\left( y+{{y}_{0}} \right)=ma$,
Now,
$\Rightarrow k{{y}_{0}}-k\left( y+{{y}_{0}} \right)=ma$,(as $mg=k{{y}_{0}}$)
On solving the above equation we get,
$a=-\left( \dfrac{k}{m} \right)y$.
Thus from this derived equation we can see that the acceleration is proportional to the displacement and the ball will perform SHM till it is in contact with the spring and moves with the spring.
That is the ball is actually maintaining a contact with the tray or indirectly with the spring.
So, we can say from the above explanation that the correct option is option A.

Note : In the equation $mg=k{{y}_{0}}$, k is the spring constant, m is the mass of the ball g is the acceleration due to gravity. Always remember a gravitational pull is always acting on the ball which is the primary reason for the compression of the spring.