Chapter 11, Problem 38P

(a)

To determine

Identify the direction of propagation of wave.

Answer to Problem 38P

Wave propagates in left direction.

Explanation of Solution

Speed of transverse wave is $10.0\text{m}/\text{s}$ and the equation of wave is $y\left(x,t\right)=A\cos \left(\omega t+kx\right)$.

The direction of wave is determined by whether the argument of cosine function is positive or not. If it is positive, means that the wave is propagating in left direction and negative value indicates the propagation in right direction. For the wave $y\left(x,t\right)=A\cos \left(\omega t+kx\right)$, argument is positive.

Therefore, the wave propagates in left direction.

(b)

To determine

The numerical values of $A,\omega$ and $k$.

Answer to Problem 38P

$\text{A}\text{is}\text{2}\text{.0}\text{mm, }\omega is1600\text{rad}/\text{s}\text{and}kis160\text{rad}/\text{s}$.

Explanation of Solution

Speed of transverse wave is $10.0\text{m}/\text{s}$ and the equation of wave is $y\left(x,t\right)=A\cos \left(\omega t+kx\right)$.

Amplitude is the maximum displacement from the mean position. It can be found directly from the diagram.

Write the value of amplitude.

$A=2.0\text{mm}$

Write the equation to find the angular frequency.

$\omega =\frac{2\pi v}{\lambda }$

Here, $\omega$ is the angular frequency, $v$ is the speed f wave, and $\lambda$ is the wavelength.

Wavelength is the distance from beginning of a crest to the end of adjacent trough. From the graph it is found that the wavelength is $4.0\text{cm}$.

Write the equation to find the wave number.

$k=\frac{2\pi }{\lambda }$ (I)

Here, $k$ is the wave number.

Conclusion:

Substitute $3.14$ for $\pi$, $10.0\text{m}/\text{s}$ for $v$, and $4.0\text{cm}$ for $\lambda$ in the equation for $\omega$.

$\begin{array}{c}\omega =\frac{2\left(3.14\right)\left(10.0\text{m}/\text{s}\right)}{4.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}\\ =1600\text{rad}/\text{s}\end{array}$

Substitute $3.14$ for $\pi$ and $4.0\text{cm}$ for $\lambda$ in the equation for $\omega$.

$\begin{array}{c}k=\frac{2\left(3.14\right)}{4.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}\\ =160\text{rad}/\text{m}\end{array}$

Therefore, $\text{A}\text{is}\text{2}\text{.0}\text{mm, }\omega is1600\text{rad}/\text{s}\text{and}kis160\text{rad}/\text{s}$.

(c)

To determine

Identify the instants (three smallest non negative cases) at which the snap of wave is taken.

Answer to Problem 38P

The snap is shot at $\text{1}\text{.0}\text{ms,}\text{5}\text{.0}\text{ms,}\text{9}\text{.0}\text{ms}$.

Explanation of Solution

Speed of transverse wave is $10.0\text{m}/\text{s}$ and the equation of wave is $y\left(x,t\right)=A\cos \left(\omega t+kx\right)$.

Time of snap shot can be identified from the instants at which maximum displacement occurs.

Write the value of cosine function at the time of maximum displacement.

$\cos \left(\omega t+kx\right)=1$

Write the possible values for $\left(\omega t+kx\right)$ to satisfy the above condition.

$\omega t+kx=2n\pi$

Here, $n$ is any integer.

Rewrite the above relation in terms of $t$.

$t=\frac{\left(2n\pi -kx\right)}{\omega }$ . (II)

It is seen that there is a peak at $3.0\text{cm}$ when the snapshot is obtained.

Rewrite equation (I) by multiplying with $x$ on both sides.

$kx=\frac{2\pi }{\lambda }x$

Substitute $4.0\text{cm}$ for $\lambda$ and $3.0\text{cm}$ for $x$ in the above equation.

$\begin{array}{c}kx=\frac{2\pi }{4.0\text{cm}}\left(3.0\text{cm}\right)\\ =\frac{3\pi }{2}\end{array}$

Rewrite equation (II) by substituting the above value of $kx$.

$t=\frac{\left(2n\pi -\frac{3\pi }{2}\right)}{\omega }$

Write the relation between the time period and $\omega$.

$\omega =\frac{2\pi }{T}$

Here, $T$ is the time period of wave.

Rewrite the previous equation for $t$ by substituting the above relation for $\omega$.

$\begin{array}{c}t=\frac{\left(2n\pi -\frac{3\pi }{2}\right)}{\left(\frac{2\pi }{T}\right)}\\ =\left(n-\frac{3}{4}\right)T\end{array}$ (III)

Write the relation between $T$ and $v$.

$T=\frac{\lambda }{v}$ (IV)

Conclusion:

Substitute $4.0\text{cm}$ for $\lambda$ and $10.0\text{m}/\text{s}$ for $v$ in equation (IV).

$\begin{array}{c}T=\frac{4.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}{10.0\text{m}/\text{s}}\\ =4.0\times {10}^{-3}\text{s}\left(\frac{1\text{ms}}{{10}^{-3}\text{s}}\right)\\ =4.0\text{ms}\end{array}$

Substitute $1$ for $n$ and $4.0\text{ms}$ for $T$ in equation (III) to find $t$.

$\begin{array}{c}t=\left(1-\frac{3}{4}\right)\left(4.0\text{ms}\right)\\ =1.0\text{ms}\end{array}$

Substitute $2$ for $n$ and $4.0\text{ms}$ for $T$ in equation (III) to find $t$.

$\begin{array}{c}t=\left(2-\frac{3}{4}\right)\left(4.0\text{ms}\right)\\ =5.0\text{ms}\end{array}$

Substitute $3$ for $n$ and $4.0\text{ms}$ for $T$ in equation (III) to find $t$.

$\begin{array}{c}t=\left(3-\frac{3}{4}\right)\left(4.0\text{ms}\right)\\ =9.0\text{ms}\end{array}$

Therefore, the snap is shot at $\text{1}\text{.0}\text{ms,}\text{5}\text{.0}\text{ms,}\text{9}\text{.0}\text{ms}$.