Question

A pendulum is made of a rod of length L and negligible mass, but capable of thermal expansion, and a weight of negligible size. (a) Show that when the temperature increases by dT, the period of the pendulum increases by a fraction αdT/2. (b) A clock controlled by a brass pendulum keeps time correctly at 10°C. If the room temperature is 30°C, does the clock run faster or slower? What is its error in seconds per day?

a) The clock runs slower with an error of αdT/2 seconds per day.
b) The clock runs faster with an error of αdT/2 seconds per day.
c) The clock remains accurate at higher temperatures.
d) The clock's accuracy depends on the specific heat of brass.

Answer 1

A pendulum's period increases with the length and the gravitational acceleration. The period of a pendulum increases with the change in temperature and the coefficient of thermal expansion. A clock controlled by a brass pendulum will run slower with an error of αdT/2 seconds per day.

Step-by-step explanation:

A pendulum is a simple harmonic oscillator that depends on its length and the acceleration due to gravity.

The period of a pendulum is given by the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

(a) When the temperature increases by dT, the length of the pendulum increases by a fraction αdT.

So, the new length of the pendulum is L(1 + αdT).

Substituting this new length into the formula, the new period of the pendulum is T(1 + ½αdT).

(b) In the case of a clock controlled by a brass pendulum, if the room temperature increases from 10°C to 30°C, the clock runs slower with an error of αdT/2 seconds per day.

The error in seconds per day can be calculated by multiplying the period of the pendulum by the change in temperature and the coefficient of thermal expansion α.