Question

Two pendulum of length 1.21 m and 1.0 m start vibrating . At some instant , the two are in the mean position in same phase . after how many vibrations of the longer pendulum , the two will be in phase ?

Answer 1

we know, time period is given by, $T=2\pi\sqrt{\frac{L}{g}}$

where T is time period, L is length of pendulum and g is acceleration due to gravity.

hence, it is clear that time period is directly proportional to square root of length.

so, $\frac{T_1}{T_2}=\sqrt{\frac{L_1}{L_2}}$

here, $L_1=1.21m$ and $L_2=1m$

so, $\frac{T_1}{T_2}=\sqrt{\frac{1.21}{1}}$

or, $\frac{T_1}{T_2}=\sqrt{\frac{121}{100}}$

or, $\frac{T_1}{T_2}=\frac{11}{10}$

or, $10T_1=11T_2$

hence it is clear that,
10 vibrations of longer pendulum = 11 vibrations of shorter pendulum.

so, after 10 vibrations of longer pendulum, the two will be in same phase.

Answer 2

Answer:

Since time period depend on the square toot of length.

⇒T2T1=L2L1

it implies T1=1.1T2

⇒10T1=11T2

It implies that  10 oscillations of first pendulum and 10 of the second they will be in phase.

Hence, the answer is 20.