Chapter 11, Problem 33P

To determine

To Find: The adjustment in the length of the pendulum.

Answer to Problem 33P

The length of the pendulum is shortened by $0.7\text{ mm}$ .

Explanation of Solution

Given info:

Length of pendulum $l=0.9930\text{ m}$

Total seconds in 1 day $\text{ = }86,400\text{ s}$

Formula used:

Write the expression for number. of cycles per day.

  $\text{Number of cycle per day = }\frac{\text{total time of 1 day in seconds}}{\text{time require to complete 1 cycle }}$

Write the expression for ratio for time.

  $\frac{T_{\text{new}}}{T_{\text{old}}}\text{ = }\frac{\text{new number of cycles/day}}{\text{total number of cycles/day}}$

Write the expression for time in terms of length and gravity.

  $T=2\pi \times \sqrt{\frac{l}{g}}$

Here,

  $T$ is time period of pendulum,

  $l$ is the length of pendulum,

  $g$ is the acceleration due to gravity.

Calculation:

Generally, an ideal clock makes $1$ cycle in exactly $2\text{s}$ , so number of cycle for an ideal clock is determined by the above formula.

  $\text{Number of cycle per day = }\frac{86400}{2}=43200\text{ cycles / day}$

After $1\text{day}$ , clock is $30\text{s}$ slow which means that they are $15$ less cycles.

Now, clock is having $43185\text{ cycles / day}$ .

  $\frac{T_{\text{new}}}{T_{\text{old}}}\text{ = }\frac{\text{new number of cycles/day}}{\text{total number of cycles/day}}\text{ = }\frac{43185}{43200}=0.999$

Time can also be written as:

  $\begin{array}{l}\frac{2\pi \times \sqrt{\frac{l_{\text{new}}}{g}}}{2\pi \times \sqrt{\frac{l_{\text{old}}}{g}}}\text{= }0.999\\ l_{\text{new}}=0.999\times l_{\text{old}}=0.999\times 0.993=0.9923\text{ m}\end{array}$

Calculate the shorten in the length.

  $\begin{array}{c}\Delta l\text{= }1-0.9923\\ =0.0007\text{ m}\\ =0.7\text{ mm}\end{array}$

Conclusion:

Thus, the length of the pendulum is shortened by $0.7\text{ mm}$ .