Chapter 18, Problem 1PQ

(a)

To determine

The difference between two waves on a string given by $y_1\left(x,t\right)=2\cos \left(3t-10x\right)$ and $y_2\left(x,t\right)=2\cos \left(3t+10x\right)$ .

Answer to Problem 1PQ

The wave 1 propagates in $+x$ direction and wave 2 propagates in $-x$ direction.

Explanation of Solution

Write the expression for a wave travelling along $+x$ direction.

$y_x\left(x,t\right)=y_{\text{max}}\sin \left(kx-\omega t+\phi \right)$        (I)

Here, $y_x\left(x,t\right)$ is the displacement of wave travelling along $+x$ direction, $k$ is the wave number, $\omega$ is the angular frequency, and $\phi$ is the initial phase.

Write the expression for a wave travelling along $-x$ direction.

$y_{-x}\left(x,t\right)=y_{\text{max}}\sin \left(kx+\omega t+\phi \right)$        (II)

Here, $y_{-x}\left(x,t\right)$ is the displacement of wave travelling along $-x$ direction.

Write the given waves in string.

$y_1\left(x,t\right)=2\cos \left(3t-10x\right)$        (III)

$y_2\left(x,t\right)=2\cos \left(3t+10x\right)$        (IV)

Conclusion:

Consider the mathematical expression $\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta$.

Rearrange equation (III) using above equation and change wave in equation (III) into sign wave.

$\begin{array}{c}{y}_1\left(x,t\right)=2\cos \left(3t-10x\right)\\ =2\sin \left(\frac{\pi }{2}-\left(3t-10x\right)\right)\\ =2\sin \left(10x-3t+\frac{\pi }{2}\right)\end{array}$        (V)

The above equation represents same cosine wave in terms of sinusoidal wave with initial phase of $\frac{\pi }{2}$.

Now comparing above equation with equation (I) it is clear that equation (V) and (I) are matching. Thus, the wave is travelling along $+x$ direction.

Consider the mathematic expression $\sin \left(\theta +\frac{\pi }{2}\right)=\cos \theta$.

Rearrange equation (IV) using above equation and change the wave in equation (IV) into sign wave.

$\begin{array}{c}{y}_1\left(x,t\right)=2\cos \left(3t+10x\right)\\ =2\sin \left(\frac{\pi }{2}+\left(3t+10x\right)\right)\\ =2\sin \left(10x+3t+\frac{\pi }{2}\right)\end{array}$        (VI)

The above equation represents same cosine wave in terms of sinusoidal wave with initial phase of $\frac{\pi }{2}$.

Now comparing above equation with equation (II) it is clear that the equations (VI) and (II) are matching. Thus, the wave is travelling along $-x$ direction.

Therefore, the wave 1 propagates in $+x$ direction and wave 2 propagates in $-x$ direction.

(b)

To determine

The wave that results on this string.

Answer to Problem 1PQ

The wave that results on this string is $4\cos 3t\cos 10x$.

Explanation of Solution

Rewrite equations (III) and (IV).

$y_1\left(x,t\right)=2\cos \left(3t-10x\right)$        (III)

$y_2\left(x,t\right)=2\cos \left(3t+10x\right)$        (IV)

When these waves moves in medium with one along $+x$ and other along $-x$, they superimpose to give standing wave.

Write the expression for the resultant of these waves.

$y\left(x,t\right)=y_1\left(x,t\right)+y_2\left(x,t\right)$        (VII)

Conclusion:

Substitute (III) and (IV) in (VII) to get resultant of two waves.

  $y\left(x,t\right)=2\cos \left(3t-10x\right)+2\cos \left(3t+10x\right)$

Use trigonometric equation, $\cos \left(A+B\right)+\cos \left(A-B\right)=2\cos A\cos B$ to expand above equation.

  $\begin{array}{c}y\left(x,t\right)=2\cos \left(3t-10x\right)+2\cos \left(3t+10x\right)\\ =2\left(\cos \left(3t-10x\right)+\cos \left(3t+10x\right)\right)\\ =2\left(\left(2\cos \left(3t\right)\cos \left(10x\right)\right)\right)\\ =4\cos \left(3t\right)\cos \left(10x\right)\end{array}$

Therefore, the wave that results on this string is $4\cos 3t\cos 10x$ .

(c)

To determine

The amplitude of resultant wave.

Answer to Problem 1PQ

The amplitude of resultant wave is $4\cos 10x$ .

Explanation of Solution

The resultant standing wave is $y\left(x\right)=4\cos \left(3t\right)\cos \left(10x\right)$.

Write the standard equation of a standing wave produced from waves given in (I) and (II).

  $y\left(x,t\right)=\left[2y_{\text{max}}\cos kx\right]\cos \omega t$

In above equation the term in square bracket $2y_{\text{max}}\sin kx$ represents the amplitude of the wave.

Conclusion:

Comparing above two equations, it is clear that $4\cos 10x$ is the amplitude of the standing wave that results from the given waves.

Therefore, the amplitude of resultant wave is $4\cos 10x$.