Chapter 16, Problem 8P

A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0 cm, a wavelength of 35.0 cm, and a frequency of 12.0 Hz. The transverse position of an element of the medium at t = 0, x = 0 is y = −3.00 cm, and the element has a positive velocity here. We wish to find an expression for the wave function describing this wave. (a) Sketch the wave at t = 0. (b) Find the angular wave number k from the wavelength. (c) Find the period T from the frequency. Find (d) the angular frequency ω and (e) the wave speed v. (f) From the information about t = 0, find the phase constant ϕ. (g) Write an expression for the wave function y(x, t).

To determine

The sinusoidal wave at time $t=0$.

Answer to Problem 8P

Therefore, the sinusoidal wave at time $t=0$ is shown in figure I.

Figure I

Explanation of Solution

The wave is traveling in the negative $x$ direction, amplitude of the wave is $20.0\text{cm}$, wavelength of the wave is $35.0\text{cm}$, frequency of the wave is $12.0\text{Hz}$ and the transverse position of an element of the medium at $t=0,x=0$ is $y=-3.0\text{cm}$

The general expression for the wave function of a sinusoidal wave is,

    $y=A\sin \left(\frac{2\pi }{\lambda }x-2\pi ft+\varphi \right)$

Here, $y$ is the transverse position of an element of the medium, $A$ is the amplitude of the wave, $\lambda$ is the wavelength, $f$ is the frequency, and $\varphi$ is the phase.

Substitute $20\text{cm}$ for $A$, $35.0\text{cm}$ for $\lambda$ and $12.0\text{Hz}$ for $f$ in the above equation.

    $\begin{array}{c}y=\left(20\text{cm}\right)\sin \left(\frac{2\pi }{35.0\text{cm}}x-2\pi \left(12{\text{s}}^{-1}\right)t+\varphi \right)\\ =\left(20\text{cm}\right)\sin \left(\left(0.18{\text{cm}}^{-1}\right)x-\left(75.36{\text{s}}^{-1}\right)t+\varphi \right)\end{array}$        (I)

At $t=0,x=0$ the value of $y=-3.0\text{cm}$ therefore,

    $\begin{array}{c}-3.0\text{cm}=\left(20\text{cm}\right)\sin \left(\left(0.18{\text{cm}}^{-1}\right)\times 0-\left(75.36{\text{s}}^{-1}\right)\times 0+\varphi \right)\\ \sin \varphi =-0.15\\ \varphi =-8.63°\\ =-0.15\text{rad}\end{array}$

Substitute $-0.15\text{rad}$ for $\varphi$ in equation (I).

    $y=\left(20\text{cm}\right)\sin \left(\left(0.18{\text{cm}}^{-1}\right)x-\left(75.36{\text{s}}^{-1}\right)t-0.15\right)$        (II)

At $t=0$, the above equation becomes,

    $y=\left(20\text{cm}\right)\sin \left(\left(0.18{\text{cm}}^{-1}\right)x-0.15\right)$        (III)

The graph for the above equation (III) is shown below.

Figure I

Conclusion:

Therefore, the sinusoidal wave at time $t=0$ is shown in figure I.

To determine

The angular wave number.

Answer to Problem 8P

The angular wave number is $0.18{\text{cm}}^{-1}$.

Explanation of Solution

Formula to calculate the angular wave number is,

    $k=\frac{2\pi }{\lambda }$

Substitute $35.0\text{cm}$ for $\lambda$ to find $k$.

    $\begin{array}{c}k=\frac{2\pi }{35.0\text{cm}}\\ =0.18{\text{cm}}^{-1}\end{array}$

Conclusion:

Therefore, the angular wave number is $0.18{\text{cm}}^{-1}$.

To determine

The period of the wave from frequency.

Answer to Problem 8P

The period of the wave from frequency is $0.083\text{s}$.

Explanation of Solution

Formula to calculate the period is,

    $T=\frac{1}{f}$

Substitute $12.0\text{Hz}$ for $f$ to find $T$.

    $\begin{array}{c}T=\frac{1}{12.0\text{Hz}}\\ =0.083\text{s}\end{array}$

Conclusion:

Therefore, the period of the wave from frequency is $0.083\text{s}$.

To determine

The angular frequency of the wave.

Answer to Problem 8P

The angular frequency of the wave is $75.36\text{rad}/\text{s}$.

Explanation of Solution

Formula to calculate the angular frequency is,

    $\omega =2\pi f$

Substitute $12.0\text{Hz}$ for $f$ to find $\omega$.

    $\begin{array}{c}\omega =2\pi \times 12.0\text{Hz}\\ =75.36\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the angular frequency of the wave is $75.36\text{rad}/\text{s}$.

To determine

The speed of the wave.

Answer to Problem 8P

The speed of the wave is $240\text{cm}/\text{s}$.

Explanation of Solution

Formula to calculate the speed of the wave is,

    $v=f\lambda$

Substitute $12.0\text{Hz}$ for $f$ and $20.0\text{cm}$ for $\lambda$ to find $v$.

    $\begin{array}{c}v=12.0\text{Hz}\times 20.0\text{cm}\\ =240\text{cm}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the wave is $240\text{cm}/\text{s}$.

To determine

The phase constant $\varphi$.

Answer to Problem 8P

The phase constant $\varphi$ is $-0.15\text{rad}$.

Explanation of Solution

As calculated in part (a), the phase constant is,

    $\varphi =-0.15\text{rad}$

Conclusion:

Therefore, the phase constant $\varphi$ is $-0.15\text{rad}$.

To determine

The expression for the wave function $y\left(x,t\right)$.

Answer to Problem 8P

The expression for the wave function is $y\left(x,t\right)=20.0\sin \left(0.18x-75.36t-0.15\right)$.

Explanation of Solution

The expression for the wave function from equation (II) is,

    $y\left(x,t\right)=\left(20\text{cm}\right)\sin \left(\left(0.18{\text{cm}}^{-1}\right)x-\left(75.36{\text{s}}^{-1}\right)t-0.15\right)$

Or it may be written as,

    $y\left(x,t\right)=20.0\sin \left(0.18x-75.36t-0.15\right)$

Here, $y$ and $x$ are in centimeters and $t$ in seconds.

Conclusion:

Therefore, the expression for the wave function $y\left(x,t\right)=20.0\sin \left(0.18x-75.36t-0.15\right)$.