Chapter 14, Problem 98A

(a)

To determine

To Explain: Adjustment needed to the pendulum to keep a better time.

Explanation of Solution

Introduction:

The period of a pendulum depends on the length of the pendulum and can be expressed as follows:

  $T=2\pi \sqrt{\frac{l}{g}}$

The clock is losing time which means that the period has increased. In order to show correct time, the length of the pendulum must be shortened which would make the period smaller and clock would run faster.

Conclusion:

Hence, to increase the speed of the clock, the length of the pendulum must be shortened.

(b)

To determine

To Calculate: Change in the length to make the period shorten by $0.0400$ s.

Answer to Problem 98A

Length of the pendulum should be shortened by $\text{1}\text{.52 cm}$ .

Explanation of Solution

Given:

Length of the pendulum, $l_1=15.0$ cm

Change in period of the pendulum, $\Delta T=0.0400$ s

Formula used:

Change in the period of the pendulum can be expressed follows:

  $\Delta T=2\pi \sqrt{\frac{l_2}{g}}-2\pi \sqrt{\frac{l_1}{g}}$

Here, l represents length and g represents acceleration due to gravity.

Calculation:

Substitute the values and solve for $l_2$ :

  $\begin{array}{l}\Delta T=2\pi \sqrt{\frac{l_1}{g}}-2\pi \sqrt{\frac{l_2}{g}}\\ 0.0400\text{s }=-2\pi \sqrt{\frac{l_2}{g}}+2\left(3.14\right)\sqrt{\frac{0.15\text{m}}{9.8{\text{m/s}}^{\text{2}}}}\\ -0.0400\text{s + 0}\text{.7769 s = }2\pi \sqrt{\frac{l_2}{g}}\\ 2\pi \sqrt{\frac{l_2}{g}}=0.7369\text{s}\\ \sqrt{\frac{l_2}{g}}=0.1173\text{s}\\ l_2=0.1348\text{m = 13}\text{.48 cm}\end{array}$

Change in length:

  $\begin{array}{l}\Delta l=l_1-l_2\\ \Delta l=15.0\text{cm -13}\text{.48 cm}\\ \Delta l\text{=1}\text{.52 cm}\end{array}$

Conclusion:

Hence, the length of the pendulum should be shortened by $\text{1}\text{.52 cm}$ .

(c)

To determine

To Calculate: The required change in length of the pendulum such that it shows correct time.

Answer to Problem 98A

The length of the pendulum must be shortened by $\text{ 5 mm}$ .

Explanation of Solution

Given:

Length of the pendulum, $l_1=77.5\text{cm = 0}\text{.775 m}$

Change in period of the pendulum, $\Delta T=5\min =300\text{s}$

Formula used:

The period of a pendulum can be expressed follows:

  $T=2\pi \sqrt{\frac{l}{g}}$

Here, l represents length and g represents acceleration due to gravity.

Calculation:

The period of the pendulum when its length is $l_1$ :

  $\begin{array}{l}{T}_1=2\pi \sqrt{\frac{l_1}{g}}\\ T_1=2\pi \sqrt{\frac{0.775\text{m}}{9.8{\text{m/s}}^{\text{2}}}}\\ T_1=1.766\text{s}\end{array}$

In 1-day 5 min, the number of oscillations of the pendulum:

  $\begin{array}{l}\frac{\text{1day}\left(\text{24 h/1day}\right)\left(\text{3600 s/1h}\right)\text{+5 min}\left(\text{60s/1min}\right)}{1.766\text{s}}\\ =49094\text{oscillations}\end{array}$

The period of the pendulum would slow by:

  $\frac{300\text{s}}{\text{49094}\text{oscillations}}=0.006\text{s/oscillations}$

The new period of the pendulum:

  $\begin{array}{c}{T}_2=1.766\text{s}-0.006\text{s}\\ \text{=1}\text{.760}\text{s}\end{array}$

Length of the pendulum, $l_2$ can be calculated as follows:

  $\begin{array}{c}{l}_2={\left(\frac{T_2}{2\pi }\right)}^{2}g\\ ={\left(\frac{1.760\text{s}}{2\pi }\right)}^2\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =0.770\text{m}\end{array}$

Change in length:

  $\begin{array}{c}\Delta l=l_1-l_2\\ =0.775\text{m}-0.770\text{m}\\ \text{= 0}\text{.005 m }\\ \text{= 5 mm}\end{array}$

Conclusion:

Hence, the length of the pendulum must be shortened by $\text{ 5 mm}$ .