Chapter 1.2, Problem 51SPR

(a)

To determine

To Find: The period of the pendulum in the clock to the nearest tenth of the second.

Answer to Problem 51SPR

The value of the period is $2.1\text{sec}$ .

Explanation of Solution

Given:

The formula for the period of the pendulum is $P=2\pi \sqrt{\frac{l}{32}}$ .

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the given formula is,

  $P=2\pi \sqrt{\frac{l}{32}}$

The length of the pendulum in ft. is,

  $\begin{array}{c}42\text{in}=42\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =3.5\text{ft}\end{array}$

Then, the period is,

  $\begin{array}{c}P=2\pi \sqrt{\frac{\text{3}\text{.5}}{32}}\\ =2.1\text{sec}\end{array}$

(b)

To determine

To Find: The length in inches long that should the pendulum be in order for it to have the period of 1 sec.

Answer to Problem 51SPR

The pendulum should have the length of $9.7\text{sec}$ .

Explanation of Solution

Given:

The length of pendulum when the period is of 1 second.

Calculation:

Consider the given formula is,

  $P=2\pi \sqrt{\frac{l}{32}}$

Then,

  $\begin{array}{l}\text{1}\sec =2\pi \sqrt{\frac{l}{32}}\\ l=32{\left(\frac{2.1\text{sec}}{2\pi }\right)}^2\\ l=\left(\frac{\sqrt{32}}{4{\pi }^2}\right)\text{ft}\\ l=\frac{32}{4{\pi }^2}\left(\frac{12\text{in}}{1\text{ft}}\right)\text{in}\end{array}$

Solve further as,

  $l=9.7\text{in}$

To obtain the period of $1\text{sec}$ the pendulum should have the length of $9.7\text{sec}$ .