Question

A cork floating on the pond water executes a simple harmonic motion, moving up and down over a range of 4 cm. The time period of the motion is 1 s. at `t=0`, the cork is at its lowest position of oscillation, the position and velocity of the cork at `t=10.5s`, would be
A. 2 cm above the mean position, `0(m)/(s)`
B. 2 cm below the mean position `0(m)/(s)`
C. 1 cm above the mean position `2sqrt3pi(m)/(s)`up
D. 1 cm below the mean position, `2sqrt3pi(m)/(s)` up

Answer 1

Correct Answer - A
As the range of motion is 4 cm the amplitude of motion is `+2cm`. 10.5 is equal to 10.5 time period of simple harmonic motion, so we have to find the height and position of cork after one half of time period at `(T)/(2)=0.5s`. As at `t=10s`, the particle is at its lowest position, after half a time period the cork would be at its maximum height and velocity of cork at extreme position is zero.