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Expert-verified Found in: Page 292 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # Your grandfather clock’s pendulum has a length of 0.9930 m. If the clock runs slow and loses 21 s per day, how should you adjust the length of the pendulum?

You should adjust the length of the pendulum by decreasing it by 0.0005 m.

See the step by step solution

## Understanding the length of the pendulum

In order to find the length required to adjust the pendulum, firstly evaluate the new length of the clock’s pendulum and then subtract it with the original value of length.

## Identification of the given information

The length of the grandfather clock’s pendulum is l = 0.9930 m.

The time lost by the clock in a day is 21 s.

The acceleration due to gravity on the Earth is, $$g = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$.

## Determination of the number of cycles made by clock per day

The number of seconds in a day is calculated as:

\begin{aligned}{l}t &= (24\;{\rm{h}})\left( {\frac{{60\;{\rm{min}}}}{{1\;{\rm{h}}}}} \right)\left( {\frac{{60\;{\rm{s}}}}{{1\;{\rm{min}}}}} \right)\\t &= 86400\;{\rm{s}}\end{aligned}

The clock completes one cycle of tick and tock in 2 seconds. Therefore, the number of cycle's clock make per day is:

\begin{aligned}{l}n &= \frac{t}{2}\\n &= \frac{{86,400}}{2}\\n &= 43,200\end{aligned}

## Determination of the new time period of the clock

The time period of oscillation of a pendulum for small amplitudes is given as:

$$T = 2\pi \sqrt {\frac{l}{g}}$$ ….. (i)

Here, g is the acceleration due to gravity and l is the length of the pendulum.

If the new length of the clock’s pendulum is $$l'$$when it runs slow, then the new time period of the pendulum is:

$$T' = 2\pi \sqrt {\frac{{l'}}{g}}$$ ….. (ii)

If the clock runs slow and loses 21 s in a day, then it means it makes 10.5 cycles less per day. Therefore, number of cycles this clock now makes per day is:

$$(43200 - 10.5)\;{\rm{cycles}} = 43189.5\;{\rm{cycles}}$$

Thus, the new time period of the clock reduces by the factor of $$\frac{{43,189.5}}{{43,200}}$$.

So, the new time period of clock is,

$$T' = \frac{{43,189.5}}{{43,200}}T$$

## Determination of the new length of the pendulum

Substitute the values of T and $$T'$$ from equation (i) and (ii) in the above expression.

\begin{aligned}{c}2\pi \sqrt {\frac{{l'}}{g}} &= \frac{{43,189.5}}{{43,200}}\left( {2\pi \sqrt {\frac{l}{g}} } \right)\\\sqrt {l'} &= \frac{{43,189.5}}{{43,200}}\sqrt l \\l' &= {\left( {\frac{{43,189.5}}{{43,200}}} \right)^2}l\end{aligned}

On substituting the value of l, you will get:

\begin{aligned}{c}l' &= {\left( {\frac{{43189.5}}{{43200}}} \right)^2} \times \left( {0.9930\;{\rm{m}}} \right)\\ &= 0.9925\;{\rm{m}}\end{aligned}

The difference between the new length and old length is:

\begin{aligned}{c}l - l' &= 0.9930\;{\rm{m}} - 0.9925\;{\rm{m}}\\ &= 0.0005\;{\rm{m}}\end{aligned}

Thus, the length of the pendulum is decreased by 0.0005 m. ### Want to see more solutions like these? 