Chapter 11, Problem 32P

(a)

To determine

To Calculate: The frequency of simple pendulum.

Answer to Problem 32P

The frequency at which it vibrates is $0.572\text{ Hz}$ .

Explanation of Solution

Given:

Length of simple pendulum $l=0.76\text{ m}$

Formula used:

Frequency of the pendulum is obtained by the following formula,

  $f=\frac{1}{2\pi }\times \sqrt{\frac{g}{l}}$

Here,

  $g$ is the acceleration due to gravity.

  $l$ is the length of simple pendulum.

Calculation:

Substitute the given values in the above equation,

  $\text{Frequency }f=\frac{1}{2\pi }\times \sqrt{\frac{9.8}{0.76}}=0.572\text{ Hz}$

Conclusion:

Thus, the frequency of pendulum at which it vibrates is $0.572\text{ Hz}$ .

(b)

To determine

To Calculate: The speed of pendulum bob.

Answer to Problem 32P

The speed of pendulum bob is $0.571\text{ m/s}$ .

Explanation of Solution

Given:

Length of pendulum $l=0.76\text{ m}$

Angle $\theta =12°$

Value of kinetic energy at top (at release point), $KE\text{ (top) }=0$

  $KE\text{ (bottom) }=\frac{1}{2}\times m\times v_{\max }{}^2$ (at lowest point KE is maximum)

Value of potential energy at top $PE\text{ (top) }=m\times g\times (L-L\cos \theta )$

Value of potential energy at bottom $PE\text{ (bottom) }=0$

Formula used:

The conservation of energy can be used and related to the lowest point to the release point of pendulum, speed of the pendulum bob is determined by the following formula,

  $E\text{(top) }=E\text{(bottom)}$

It can be written as

  $KE\text{(top) }+PE\text{(top) }=KE\text{(bottom) }+PE\text{(bottom)}$

Here $KE$ represent the kinetic energy of the bob and $PE$ represent the potential energy of the bob.

Calculation:

Substitute the given values,

  $\begin{array}{l}KE(top)+PE\text{(top) }=KE\text{(bottom) }+PE\text{(bottom)}\\ 0+mg(L-L\cos \theta )=\frac{1}{2}\times m\times v_{\max }{}^2+0\\ v_{\max }{}^2=2\times g\times (L-L\cos \theta )=2\times 9.8\times 0.76\times (1-\cos 12°)\\ v_{\max }=0.571\text{ m/s}\end{array}$

The speed of pendulum bob is $0.571\text{ m/s}$ .

Conclusion:

Thus, the speed of pendulum bob is $0.571\text{ m/s}$ .

(c)

To determine

To Calculate: The total energy stored in oscillation.

Answer to Problem 32P

The total energy stored in this oscillation is $5.95\times {10}^{-2}\text{ J}$ .

Explanation of Solution

Given:

Maximum velocity $v_{\max }=0.571\text{ m/s}$

Mass $m=0.365\text{ kg}$

Formula used:

The total energy can be found from the KE at the bottom of motion by the following formula,

  $E_{\text{total}}=\frac{1}{2}\times m\times v_{\max }{}^2$

Here,

  $m$ is mass of pendulum.

  $v_{\max }$ is maximum speed of bob.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{E}_{\text{total}}=\frac{1}{2}\times 0.365\times {0.571}^2\\ E_{\text{total}}=5.95\times {10}^{-2}\text{ J}\end{array}$

Conclusion:

Thus, the total energy stored in this oscillation is $5.95\times {10}^{-2}\text{ J}$ .