Question

Two systems are in oscillation: a simple pendulum swinging back and forth through a very small angle and a block oscillating on a spring. The block-spring system takes twice as much time as the pendulum to complete one oscillation. Which of the following changes could make the two systems oscillate with the same period?

A. Increasing the mass of the pendulum bob

B. Increasing the angle through which the pendulum swings by a small amount

C. Decreasing the mass of both the block and the pendulum bob

D. Shortening the pendulum

Answer 1

Given Two oscillating systems are
a simple pendulum swinging back and forth through a very small angle

and a block oscillating on a spring.

we know that the time period of the two systems are

for simple pendulum T1 = 2pi sqrt(l/g)

and for spring block system T2 = 2pi sqrt(m/k)

from the above expressions T1 does not depend on mass of the bob it only depens on length of the pendulum and g value at that location

and T2 dpends on mass of the block and the spring constant , if the spring constnat , remains same , only it depends on mass of the block

so also given that T2 = 2T1

to make T2 = T1, means to decrease the time period of the spring mass system , we should decrease the mass of the block, only

by decreasing the bod mass there is no change in the time period (T1)

Explanation

A. Increasing the mass of the pendulum bob, makes no change in the time period of the simple pendulum

B.Increasing the angle through which the pendulum swings by a small amount , so not a correct option

here the angle is in small amount so there is no much change in timeperiod of the pendulum so this is also not a correct option

C. Decreasing the mass of both the block and the pendulum bob, mass of the block decreases the time period decreases , it is only the correct option

D. Shortening the pendulum, makes the time period of the pendulum still decreases, which does not satisfices the condition of T1 = T2