First slide

Simple hormonic motion

Question

Two pendulum of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is $\frac{5 T}{4}$ , where T is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.

Moderate

A

1 : 16
B

1 : 4
C

1 : 2
D

1 : 25

Solution

$Let T_{1} = T and T_{2} = KT, \frac{2 π}{T} t - \frac{2 π}{KT} t = 2 π$
$t (\frac{K - 1}{KT}) = 1; \frac{5 T}{4} (\frac{K - 1}{KT}) = 1 which gives K = 5$
If length of first pendulum = l, $T_{1} = 2 π \sqrt{\frac{l}{g}}$
$T_{2} = 5 \times 2 π \sqrt{\frac{l}{g}} = 2 π \sqrt{\frac{25 l}{g}}$
So, ratio of lengths = 1 : 25

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App