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Question: There are two pendulums of length \({L_1}\) and \({L_2}\) which start vibrating. At some instant, th...

There are two pendulums of length L1{L_1} and L2{L_2} which start vibrating. At some instant, the both pendulums are in the same position in the same phase. After how many vibrations of the shorter pendulum, will both pendulums be in phase in the mean position? [(L1>L2),L1=121andL2=100][({L_1} > {L_2}),{L_1} = 121and{L_2} = 100]
A. 1111
B. 1010
C. 99
D. 88

Explanation

Solution

Simple pendulum is an instrument which has a mass attached to one end of a massless rod known as the bob of the simple pendulum and is fixed to the rigid support at the other end of the rod. When the bob is displaced slightly from the mean position, it starts oscillating with the same displacement about the mean position.

Formula used:
The time period of the simple pendulum is given by:
T=2πLgT = 2\pi \sqrt {\dfrac{L}{g}}
Where, LL is the length of the rod and gg is the acceleration due to gravity.

Complete step by step answer:
Now,
T=2πLgT = 2\pi \sqrt {\dfrac{L}{g}}
Also, let T1{T_1} and T2{T_2} be the time period of the greater length pendulum and smaller length pendulum.So,
T1T2=2πL1g2πL2g ⇒T1T2=L1L2 \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{2\pi \sqrt {\dfrac{{{L_1}}}{g}} }}{{2\pi \sqrt {\dfrac{{{L_2}}}{g}} }} \\\ \Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{\sqrt {{L_1}} }}{{\sqrt {{L_2}} }} \\\
Now putting the values of L1L_1 and L2L_2,
T1T2=121100 ⇒T1T2=1110 \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{\sqrt {121} }}{{\sqrt {100} }} \\\ \Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{11}}{{10}} \\\
∴10 T1=11 T2\therefore 10\,{T_1} = 11\,{T_2}
According to this, after 11 vibration of the shorter pendulum both pendulums will be in the same phase.

So, option A is the correct answer.

Note: We know that the bob of a simple pendulum moves to and fro about the mean position. So, according to the relation of the time period the pendulum that has the smaller length, it finishes the to and fro motion faster as compared to that of the longer length pendulum. So, the time period of the smaller length pendulum is less than that of the longer length pendulum. That is why it requires 10 vibrations of longer length and 11 vibration of smaller length pendulum.