Chapter 11, Problem 21P

(a)

To determine

The amplitude of the motion.

Answer to Problem 21P

The amplitude of motion is $0.38\text{ m}$ .

Explanation of Solution

Given:

Mass, $m=0.3\text{ kg}$

  $x=0.38\sin (6.5t)$

Formula used:

Compare the given equation with $x=A\sin (\omega t)$ , to find the value of amplitude A

Calculation:

By comparing these two equations, amplitude A in meter is obtained by,

  $\begin{array}{l}x=0.38\sin (6.5t)\\ x=A\sin (\omega t)\end{array}$

By comparing the equation:

Amplitude, $A=0.38\text{ m}$

Conclusion:

Thus, the amplitude of motion is $0.38\text{ m}$ .

(b)

To determine

The frequency of the wave.

Answer to Problem 21P

The frequency is $1.03\text{ Hz}$ .

Explanation of Solution

Given:

  $x=0.38\sin (6.5t)$

Angular speed, $\omega =6.5\text{ rad/s}$

Formula used:

The relationship between the angular speed and frequency is,

Angular speed, $\omega =2\times \pi \times f$

Calculation:

From the above equation the frequency is determined by following way,

  $\begin{array}{l}f=\frac{\omega }{2\times \pi }\\ f=\frac{6.5}{2\times \pi }=1.03\text{ Hz}\end{array}$

Conclusion:

Thus, the frequency is $1.03\text{ Hz}$

(c)

To determine

The period of the cycle.

Answer to Problem 21P

The period is $0.97\text{s}$ .

Explanation of Solution

Given:

Frequency, $f=1.03\text{ Hz}$

Formula used:

The relationship between the time period and frequency is,

  $T=\frac{1}{f}$

Calculation:

Substitute the available value of frequency in the above equation,

Period, $T=\frac{1}{1.03}=0.97\text{ s}$

Conclusion:

Thus, the period is $0.97\text{ s}$ .

(d)

To determine

The total energy of the system.

Answer to Problem 21P

The total energy of the system is $0.92\text{ J}$ .

Explanation of Solution

Given:

Mass, $m=0.3\text{ kg}$

Angular speed, $\omega =6.5\text{ rad/s}$

Amplitude, $A=0.38\text{ m}$

Formula used:

The total energy of the system is determined by the following formula,

  $\begin{array}{l}{E}_{total}=\frac{1}{2}\times m\times v^2\\ v=\omega \times A\\ \text{so, }{E}_{total}=\frac{1}{2}\times m\times {(}^{\omega }\end{array}$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{E}_{total}=\frac{1}{2}\times 0.30\times {(}^{6.5}\\ E_{total}=0.92\text{ J}\end{array}$

Conclusion:

Thus, the total energy of the system is $0.92\text{ J}$ .

(e)

To determine

To Calculate: The kinetic energy and potential energy.

Answer to Problem 21P

The kinetic energy at x = 0.09 m is $0.86\text{ J}$ & the potential energy at x = 0.09 m is $0.051\text{ J}$ .

Explanation of Solution

Given:

Mass, $m=0.3\text{ kg}$

Angular speed, $\omega =6.5\text{ rad/s}$

Amplitude, $A=0.38\text{ m}$

Distance, $x=0.09\text{ m}$

Formula used:

The kinetic energy is calculated by the following formula.

  $KE=\frac{1}{2}\times m\times {\omega }^2\times (A^2-x^2)$

The potential energy is calculated by the following formula,

  $PE=\frac{1}{2}\times m\times {\omega }^2\times x^2$Here $m$ is the mass,

  $\omega$ is the angular speed.

  $A$ is the amplitude,

  $x$ is the distance.

Calculation:

Substitute the given values in the above equation in the above equation,

  $\begin{array}{l}KE=\frac{1}{2}\times 0.3\times {6.5}^2\times ({0.38}^2\text{- }{0.09}^2)\\ KE=0.86\text{ J}\end{array}$

  $\begin{array}{l}PE=\frac{1}{2}\times 0.3\times {6.5}^2\times {0.09}^2\\ PE=0.051\text{ J}\end{array}$

Conclusion:

Thus, the kinetic energy at x = 0.09 m is $0.86\text{ J}$ & the potential energy at x = 0.09 m is $0.051\text{ J}$ .

(f)

To determine

To Draw: The graph of x versus t .

Explanation of Solution

Graph:

  

Conclusion:

The graph shows a sin relation between the time and distance with a cycle. The maximum amplitude achieved is 0.38 m and the time period for each complete cycle is 0.97 s.