Chapter 13, Problem 18P

The pendulum in a grandfather clock is made of brass and keeps perfect time at 17°C. How much time is gained or lost in a year if the clock is kept at 29°C? (Assume the frequency dependence on length for a simple pendulum applies.) [Hint:See Chapter 8)

To determine

The time lost or gained in a year if the pendulum clock is kept at the given higher temperature

Answer to Problem 18P

Solution:

The pendulum clock will lose approximately 60 minutes (or 1 hour)in a year.

Explanation of Solution

Given:

Initial temperature, $T_i=17°C$

Final temperature, $T_f{}_{}=29°C$

The pendulum is made of brass

Formula used:

  i)                  The time period of a pendulum clock is given by,

                    $t_o=2\pi \sqrt{\frac{l_0}{g}}$, where l₀ is the length of the pendulum

  ii)                The relation between change in length of a substance and change in temperature is given by,

                    $\Delta l=l_0\times \alpha \times \Delta T$, where ΔT is the change in temperature andΔ is the coefficient of linear expansion of a material.

Calculation:

The time period of a pendulum clock is given by,

$t_o=2\pi \sqrt{\frac{l_0}{g}}$, where l₀ is the length of the pendulum

Since the time period depends on the length of the pendulum, a change in this length because of change in temperature will have an effect on the time period also

The relation between change in length of a substance and change in temperature is given by,

$\Delta l=l_0\times \alpha \times \Delta T$, where ΔT is the change in temperature andΔ is the coefficient of linear expansion of a material.

If Δt is the time lost or gained, we have

$\frac{\Delta t}{t_0}=\frac{t_f-t_0}{t_0}$

      = $\frac{2\pi \sqrt{\frac{l_f}{g}}-2\pi \sqrt{\frac{l_0}{g}}}{2\pi \sqrt{\frac{l_0}{g}}}$

      = $\frac{\sqrt{l_f}-\sqrt{l_0}}{\sqrt{l_0}}$

      = $\frac{\sqrt{l_0+\alpha l_0\Delta T}-\sqrt{l_0}}{\sqrt{l_0}}$

or $\frac{\Delta t}{t_0}=\sqrt{1+\alpha \Delta T}-1$

or $\Delta t=\left[\sqrt{1+\alpha \Delta T}-1\right]\left(t_o\right)$

For brass, $\alpha =19\times {10}^{-6}\frac{1}{°C}$

Based on the given data,$\Delta T=T_f–T_i=29°C-17°C=12°C$

$t=1year=365\times 24\times 60\times 60s=3.15\times {10}^{7}s$

Therefore, $\Delta t=\left[\sqrt{1+(19\times {10}^{-6}\frac{1}{°C})(12°C)}-1\right]\left(3.15\times {10}^{7}s\right)$

$=3591s$

$=60min$