Chapter 16, Problem 77P

(a)

To determine

Wave function of two waves.

Explanation of Solution

Given:

The wave function of standing wave is given as $y\left(x,t\right)=\left(0.020\right)\sin \left(\frac{1}{2}\pi x\right)\cos \left(40\pi t\right)$ .

Introduction:

The superimposed traveling waves have same wave number and same angular frequency. However, the amplitude of superimposed waves is equal and half the amplitude of the standing wave.

Write expression for the wave function for the wave traveling in positive x-direction.

  $y_1\left(x,t\right)=\left(0.010\text{m}\right)\sin \left[\left(\frac{\pi }{2}{\text{m}}^{\text{-1}}\right)x-\left(40\pi {\text{s}}^{\text{-1}}\right)t\right]$

Write expression for the wave function for the wave traveling in negative x-direction.

  $y_2\left(x,t\right)=\left(0.010\text{m}\right)\sin \left[\left(\frac{\pi }{2}{\text{m}}^{\text{-1}}\right)x+\left(40\pi {\text{s}}^{\text{-1}}\right)t\right]$

Conclusion:

Thus, the wave function for superimposed waves in positive and negative x-direction is given above.

(b)

To determine

The distance between nodes of the standing wave.

Explanation of Solution

Given:

The wave function of standing wave is given as $y\left(x,t\right)=\left(0.020\right)\sin \left(\frac{1}{2}\pi x\right)\cos \left(40\pi t\right)$ .

Formula used:

Write expression for wave number.

  $k=\frac{2\pi }{\lambda }$

Rearrange above expression for $\lambda$ .

  $\lambda =\frac{2\pi }{k}$

Write expression for distance between adjacent nodes.

  $d=\frac{1}{2}\lambda$

Substitute $\frac{2\pi }{k}$ for $\lambda$ in above expression.

  $d=\frac{\pi }{k}$........ (1)

Calculation:

Substitute $\frac{\pi }{2}$ for $k$ in equation (1).

  $\begin{array}{l}d=\frac{\pi }{\frac{\pi }{2}}\\ d=2\text{m}\end{array}$

Conclusion:

Thus, the distance between two waves is $2\text{m}$ .

(c)

To determine

The maximum speed of the rope at $x=1.0\text{m}$ .

Explanation of Solution

Given:

The wave function of standing wave is given as $y\left(x,t\right)=\left(0.020\right)\sin \left(\frac{1}{2}\pi x\right)\cos \left(40\pi t\right)$ .

Formula used:

Write expression for standing wave.

  $y\left(x,t\right)=\left(0.020\right)\sin \left(\frac{1}{2}\pi x\right)\cos \left(40\pi t\right)$....... (1)

Calculation:

Differentiate equation (1) with respect to $t$ .

  $\begin{array}{c}\frac{d}{dt}y\left(x,t\right)=\frac{d}{dt}\left[\left(0.020\right)\sin \left(\frac{1}{2}\pi x\right)\cos \left(40\pi t\right)\right]\\ v\left(x,t\right)=-\left(0.80\pi \right)\sin \left(\frac{\pi }{2}x\right)\sin \left(40\pi t\right)\end{array}$

Substitute $1.0\text{m}$ for $x$ in above expression.

  $\begin{array}{c}v\left(\left(1.0\text{m}\right),t\right)=-\left(0.80\pi \right)\sin \left(\frac{\pi }{2}\left(1.0\text{m}\right)\right)\sin \left(40\pi t\right)\\ v\left(\left(1.0\text{m}\right),t\right)=\left(0.80\pi \text{m}/\text{s}\right)\sin \left(40\pi {\text{s}}^{\text{-1}}\right)t\\ v\left(\left(1.0\text{m}\right),t\right)=\left(2.5\text{m}/\text{s}\right)\sin \left(40\pi {\text{s}}^{\text{-1}}\right)t\end{array}$

Substitute $v_{\max }$ for $v\left(\left(1.0\text{m}\right),t\right)$ and 1 for $\sin \left(40\pi {\text{s}}^{\text{-1}}\right)t$ in above expression.

  $v_{\max }=2.5\text{m}/\text{s}$

Conclusion:

Thus, the maximum speed of the wave at $x=1.0\text{m}$ is $2.5\text{m}/\text{s}$ .

(d)

To determine

The maximum acceleration of the rope at $x=1.0\text{m}$ .

Explanation of Solution

Given:

The wave function of standing wave is given as $y\left(x,t\right)=\left(0.020\right)\sin \left(\frac{1}{2}\pi x\right)\cos \left(40\pi t\right)$ .

Formula used:

Write expression for standing wave.

  $y\left(x,t\right)=\left(0.020\right)\sin \left(\frac{1}{2}\pi x\right)\cos \left(40\pi t\right)$....... (1)

Calculation:

Differentiate equation (1) with respect to $t$ .

  $\begin{array}{c}\frac{d}{dt}y\left(x,t\right)=\frac{d}{dt}\left[\left(0.020\right)\sin \left(\frac{1}{2}\pi x\right)\cos \left(40\pi t\right)\right]\\ v\left(x,t\right)=-\left(0.80\pi \right)\sin \left(\frac{\pi }{2}x\right)\sin \left(40\pi t\right)\end{array}$

Differentiate above expression with respect to $t$ .

  $\begin{array}{c}\frac{d}{dt}v\left(x,t\right)=\frac{d}{dt}\left[-\left(0.80\pi \right)\sin \left(\frac{\pi }{2}x\right)\sin \left(40\pi t\right)\right]\\ a\left(x,t\right)=\pm \left(32{\pi }^2\text{m}/{\text{s}}^{\text{2}}\right)\cos \left(40\pi {\text{s}}^{\text{-1}}\right)t\end{array}$

Substitute $a_{\max }$ for $a\left(x,t\right)$ and 1 for $\cos \left(40\pi {\text{s}}^{\text{-1}}\right)t$ in above expression.

  $a_{\max }=\pm 316\text{m}/{\text{s}}^{\text{2}}$

Conclusion:

Thus, maximum acceleration at $x=1.0\text{m}$ is $\pm 316\text{m}/{\text{s}}^{\text{2}}$ .