First slide

Applications of SHM

Question

Two pendulums 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 5T/4, where T is the time period of shorter pendulum. Find the ratios of lengths of the two Pendulums.

Moderate

A

1:16
B

1:4
C

1:2
D

1:25

Solution

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

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