Question

a) A stone is suspended from one end of a thread and execute a simple harmonic motion with a frequency of 0.5 Hz. If the length of the thread is increased by four times the initial length, then determine the period of the harmonic motion. 

Answer 1

The period of the harmonic motion is given as follows:

$$T = 2\pi \sqrt{\dfrac{l}{g}}$$

where $l$ is the length of the thread, and $g$ is the gravitational acceleration.

On the other hand, by definition, the period is the inverse frequency:

$$T = \dfrac{1}{f} = \dfrac{1}{0.5Hz} = 2s$$

If we increase the length by four times: $l_1= 4l$, then the period increases by 2 times:

$$T_1 =2\pi \sqrt{\dfrac{l_1}{g}} = 2\pi \sqrt{\dfrac{4l}{g}} = 2\cdot 2\pi \sqrt{\dfrac{l}{g}} = 2T$$

Since $T = 2s$, the nes period is $T_1 = 2\cdot 2s = 4s$.

Answer. 4s.