Question

The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgl}}$ where m = mass of the pendulum, I = moment of inertia about the axis of suspension, = distance of centre of mass of body from the axis of suspension. The change in time period when temperature changes by $\Delta \theta$ is given by $\frac{\alpha \Delta \theta}{n}T$, then the value of n is ____. [The coefficient of linear expansion of the material of pendulum is $\alpha$]

Answer 1

2

Answer 2

The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgl}}$. When the temperature changes by $\Delta \theta$, the length $l$ changes to $l' = l(1 + \alpha \Delta \theta)$, and the moment of inertia $I$ changes to $I' = I(1 + \alpha \Delta \theta)^2$, assuming all linear dimensions scale uniformly. The new time period is $T' = 2\pi \sqrt{\frac{I'}{mgl'}} = T \sqrt{1 + \alpha \Delta \theta}$. Using the binomial approximation for small $\alpha \Delta \theta$, $T' \approx T(1 + \frac{1}{2} \alpha \Delta \theta)$. The change in time period is $\Delta T = T' - T \approx \frac{1}{2} \alpha \Delta \theta T$. Comparing this with the given change $\frac{\alpha \Delta \theta}{n}T$, we find $\frac{1}{n} = \frac{1}{2}$, which gives $n=2$.