Chapter 17, Problem 68PQ

(a)

To determine

The wave function for the wave if $y\left(x,t\right)=4.50\text{ cm}$ at $x=0$ and $t=0$ .

Answer to Problem 68PQ

The wave function for the wave is $y\left(x,t\right)=\left(0.0450\right)\sin \left(20.0\pi x+3.00\pi t+\frac{\pi }{2}\right)$ .

Explanation of Solution

It is given that the sinusoidal wave is travelling in the negative direction.

Write the general equation for the sinusoidal wave travelling in negative $x$ direction.

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

Here, $y\left(x,t\right)$ is the displacement of wave at any position $x$ and time $t$, $y_{\text{max}}$ is the amplitude of wave, $\omega$ is the angular frequency, $k$ is the angular wavenumber and $\phi$ is the initial phase.

Write the expression for $k$ .

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

Here, $\lambda$ is the wavelength.

Write the expression for $\omega$ .

  $\omega =2\pi f$                                                                                                                (III)

Here, $f$ is the frequency of the wave.

Conclusion:

It is given that the amplitude of the wave is $4.50\text{ cm}$, the wavelength is $10.0\text{ cm}$ and the frequency is $1.50\text{ Hz}$ .

Substitute $10.0\text{ cm}$ for $\lambda$ in equation (II) to get $k$ .

  $\begin{array}{c}k=\frac{2\pi }{\left(10.0\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}\\ =\frac{2\pi }{0.100\text{ m}}\\ =20.0\pi {\text{m}}^{-1}\end{array}$

Substitute $1.50\text{ Hz}$ for $f$ in equation (III) to get $\omega$ .

  $\begin{array}{c}\omega =2\pi \left(1.50\text{ Hz}\right)\\ =3.00\pi \text{ rad/s}\end{array}$

It is given that the value of the wave function at $x=0$ and $t=0$ is $4.50\text{ cm}$ .

Substitute $4.50\text{ cm}$ for $y\left(x,t\right)$, $4.50\text{ cm}$ for $y_{\text{max}}$, $0$ for $x$ and $0$ for $t$ in equation (I) and rewrite the equation to find the value of $\phi$ .

  $\begin{array}{c}4.50\text{ cm}=\left(4.50\text{ cm}\right)\sin \left(0+0+\phi \right)\\ \sin \phi =1\\ \phi ={\sin }^{-1}\left(1\right)\\ =\frac{\pi }{2}\end{array}$

Substitute $4.50\text{ cm}$ for $y_{\text{max}}$, $20.0\pi {\text{m}}^{-1}$ for $k$, $3.00\pi \text{ rad/s}$ for $\omega$ and $\frac{\pi }{2}$ for $\phi$ in equation (I) to find $y\left(x,t\right)$ .

  $\begin{array}{c}y\left(x,t\right)=\left(4.50\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)\sin \left(\left(20.0\pi {\text{m}}^{-1}\right)x+\left(3.00\pi \text{ rad/s}\right)t+\frac{\pi }{2}\right)\\ =\left(0.0450\right)\sin \left(20.0\pi x+3.00\pi t+\frac{\pi }{2}\right)\end{array}$

Therefore, the wave function for the wave is $y\left(x,t\right)=\left(0.0450\right)\sin \left(20.0\pi x+3.00\pi t+\frac{\pi }{2}\right)$ .

(b)

To determine

The wave function for the wave if $y\left(x,t\right)=4.50\text{ cm}$ at $x=5.00\text{ cm}$ and $t=0$ .

Answer to Problem 68PQ

The wave function for the wave if $y\left(x,t\right)=4.50\text{ cm}$ at $x=5.00\text{ cm}$ and $t=0$ is $y\left(x,t\right)=\left(0.0450\right)\sin \left(20.0\pi x+3.00\pi t-\frac{\pi }{2}\right)$ .

Explanation of Solution

All quantities of the wave are same as in part (a) except for the phase. Equation (I) can be used to find the wave function.

Conclusion:

Substitute $4.50\text{ cm}$ for $y\left(x,t\right)$, $4.50\text{ cm}$ for $y_{\text{max}}$, $20.0\pi {\text{m}}^{-1}$ for $k$, $5.00\text{ cm}$ for $x$ and $0$ for $t$ in equation (I) and rewrite the equation to find the value of $\phi$ .

  $\begin{array}{c}\left(4.50\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)=\left(4.50\text{ cm}\right)\sin \left(\left(20.0\pi {\text{m}}^{-1}\right)\left(5.00\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)+0+\phi \right)\\ 0.0450=0.0450\sin \left(20.0\left(0.0500\right)\pi +\phi \right)\\ \sin \left(\pi +\phi \right)=1\\ \pi +\phi =\frac{\pi }{2}\\ \phi =-\frac{\pi }{2}\end{array}$

Substitute $4.50\text{ cm}$ for $y_{\text{max}}$, $20.0\pi {\text{m}}^{-1}$ for $k$, $3.00\pi \text{ rad/s}$ for $\omega$ and $-\frac{\pi }{2}$ for $\phi$ in equation (I) to find $y\left(x,t\right)$ .

  $\begin{array}{c}y\left(x,t\right)=\left(4.50\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)\sin \left(\left(20.0\pi {\text{m}}^{-1}\right)x+\left(3.00\pi \text{ rad/s}\right)t-\frac{\pi }{2}\right)\\ =\left(0.0450\right)\sin \left(20.0\pi x+3.00\pi t-\frac{\pi }{2}\right)\end{array}$

Therefore, the wave function for the wave if $y\left(x,t\right)=4.50\text{ cm}$ at $x=5.00\text{ cm}$ and $t=0$ is $y\left(x,t\right)=\left(0.0450\right)\sin \left(20.0\pi x+3.00\pi t-\frac{\pi }{2}\right)$ .