Question

A pendulum clock with a brass suspension system is calibrated so that its period is 1 s at 20 degree C. If the temperature increases to 43 degree C, by how much does the period change? Answer in units of s.

Answer 1

Answer:

0.207 ms

Explanation:

First of all we need to find the length of the pendulum at 20 degrees. We know that the period is 1 s, and the formula for the period is

$T=2\pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum and g is the gravitational acceleration. Solving the equation for L and using T = 1 s and g = 9.8 m/s^2, we find

$L=g(\frac{T}{2\pi})^2=(9.8) (\frac{1}{2\pi})^2=0.248237 m$

Now we can find the new length of the pendulum at 43 degrees; the coefficient of thermal expansion of brass is

$\alpha =18\cdot 10^{-6} 1/^{\circ}C$

And the new length of the pendulum is given by

$L' = L (1+\alpha \Delta T)$

where in this case

$\Delta T = 43-20 = 23^{\circ}$ is the change in temperature

Substituting,

$L'=(0.248237)(1+(18\cdot 10^{-6})(23))=0.248340 m$

So we can now calculate the new period of the pendulum:

$T'=2\pi \sqrt{\frac{L'}{g}}=2\pi \sqrt{\frac{0.248340}{9.8}}=1.000208 s$

So the change in the period is

$T'-T=1.000208 - 1.000000 = 0.000207 s = 0.207 ms$