Question

Two bobs of masses 1 kg and 2kg are suspended from a rigid support by threads of length 4m and 1m respectively. Find the ratio of their time period.
options -
1)4
2) 8
3) 2
4)12

Answer 1

Since the time period is proportional to √length and independent of mass, the ratio is 2.

Answer 2

Given:

• Masses are 1 kg and 2 kg
• Thread length are 4 m and 1 m.

To find:

• Ratio of time period of pendulum?

Calculation:

The general expression of time period for a simple pendulum is:

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

$\implies T \propto \sqrt{l }$

• Now, taking appropriate ratios, we get:

$\implies \: \dfrac{ T_{1}}{ T_{2}} = \dfrac{ \sqrt{l1} }{ \sqrt{l2} }$

$\implies \: \dfrac{ T_{1}}{ T_{2}} = \dfrac{ \sqrt{4} }{ \sqrt{1} }$

$\implies \: \dfrac{ T_{1}}{ T_{2}} = \dfrac{ 2 }{1}$

$\implies \: T_{1} : T_{2} = 2 : 1$

So, the ratio of time period is 2 : 1 .