Question

A long, narrow steel rod of length 2.5000 m at 25°C is oscillating as a pendulum about a horizontal axis through one end. If the temperature changes to 0°C, what will be the fractional change in its period?

Answer 1

Initially time period for a long narrow steel rod is `T=2pi sqrt(L/g)`

now,on decreasing temperature by `25^@C`,change in its length `deltaL=L alpha deltat =2.5*alpha*25` (where, `alpha` is the coefficient of linear thermal expansion,for steel it is `11*10^-6` units,given `L=2.5m`)

So,`delta L=687.5*10^-6 m`

So, we have the generalised equation as `T= 2pi sqrt(L/g)`

or, `(delta T)/T = 1/2 (delta L)/L` (for smaller changes)

We have, `delta L=687.5*10^-6`

so, `(delta L)/L=(687.5*10^-6)/2.5=275*10^-6`

Hence, fractional change in time period, `(delta T)/T=1/2 *275*10^-6=137.5*10^-6`