Question

A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum is 3.07 s. The temperature rises by 142 C°, and the length of the wire increases. Determine the change in the period of the heated pendulum.

Answer 1

T = 2pi* sqrt( Lo/g)

3.07 = 2pi * sqrt( Lo/9.8)

from here we got original lenght Lo = 2.33961 m

when The temperature rises by 142 C new lenght by thermal expension

L = Lo* ( 1 + alpha* T )

for  brass alpha = 19*10^-6 ( standard value )  

L = 2.33961 * ( 1+ 19.10^-6 * 142 )

L = 2.3459 m

so new time period = T = 2pi* sqrt( 2.3459/ 9.8) = 3.07413 sec

change in the period = 3.07413 -3.07 =0.00413 sec answer

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