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11-31P

Expert-verifiedFound in: Page 292

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.

** **

**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.**

** **

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, \(g = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

** **

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:

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

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,

\(T' = \frac{{43,189.5}}{{43,200}}T\)

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.

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