A pendulum clock is designed to tick off one second on each side-to-side swing o

Question

A pendulum clock is designed to tick off one second on each side-to-side swing of the pendulum (two ticks per complete period).

1. A particular pendulum clock keeps correct time at 20.0 C. The pendulum shaft is steel, and its mass can be ignored compared with that of the bob. What is the fractional change in the length of the shaft when it is cooled to 12.0 C ?

2. How many seconds per day will the clock gain or lose at 12.0 C ?

3. How closely must the temperature be controlled if the clock is not to gain or lose more than 1.00 s a day?

Explanation / Answer

part1

we know that

fraction change in length = delta L/L = alpha*delta T

alpha = linear thermal expansion coefficient of steel = 13 * 10 ^(-6) /degree C

deltaL/L = 13*10^(-6)*(20-12) = 10.4*10^(-5)

part (2).

T = time period = 2*pi*sqrt(L / g)

initially To = 2*pi*sqrt( Lo /g ) = 2 s. (because given that a period is complete in two ticks)

then T' = 2*pi*sqrt [Lo(1 - 10.4*10^(-5)) /g]

since the time period will decrease (T' < To), the clock will gain time

T' / T = sqrt(1 - 10.4*10^(-5))

T' = 0.999947 * T0

per 2 s, the clock will gain is

T' = (1 - 0.999947)*2

T' = 0.000106 s

therefore in 24 * 60 * 60 s it will gain [24 * 60 * 60/2] * 0.000106 = 4.57 s

part(3)

we know that

change in temperature

delta theta = 2*delta T / alpha*T

delta theta = 2*(+/- 1) / 13*10^(-6)*24*60*60

delta theta = +/- 1.9 degree C

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