Chapter 13, Problem 7P

(a)

To determine

The expression for $y$ as a function of $x$ and $t$ for the given sinusoidal wave.

Answer to Problem 7P

The expression for $y$ as a function of $x$ and $t$ for the given sinusoidal wave is $y\left(x,t\right)=0.08\sin \left(2.5\pi x+6.0\pi t\right)$.

Explanation of Solution

Given information:

The value of $A$ is $8.0\text{cm}$, the value of $\lambda$ is $80.0\text{cm}$, the value of $f$ is $3.0\text{Hz}$ and $y\left(0,t\right)$ is $0$ at $t=0$.

The general expression for the sinusoidal wave is,

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

$y\left(x,t\right)$ 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.

$\varphi$ is the phase.

Substitute $8.0\text{cm}$ for $A$, $80.0\text{cm}$ for $\lambda$ and $3.0\text{Hz}$ for $f$ in the above equation.

$\begin{array}{c}y\left(x,t\right)=\left(8.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\sin \left(\frac{2\pi }{80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}x+2\pi \left(3.0\text{Hz}\right)t+\varphi \right)\\ =\left(0.08\text{m}\right)\sin \left(2.5\pi {\text{m}}^{-1}x+\left(6.0\pi {\text{s}}^{-1}\right)t+\varphi \right)\end{array}$ (II)

Apply the condition $y\left(0,t\right)=0$ at $t=0$ in equation (II).

$\begin{array}{c}0=\left(0.08\text{m}\right)\sin \left(2.5\pi {\text{m}}^{-1}\times 0+6.0\pi \times 0+\varphi \right)\\ 0=\sin \varphi \\ \varphi =0\end{array}$

Substitute $0$ for $\varphi$ in equation (II).

$\begin{array}{c}y\left(x,t\right)=\left(0.08\text{m}\right)\sin \left(2.5\pi {\text{m}}^{-1}x+\left(6.0\pi {\text{s}}^{-1}\right)t\right)\\ =0.08\sin \left(2.5\pi x+6.0\pi t\right)\end{array}$

Conclusion:

Therefore, the expression for $y$ as a function of $x$ and $t$ for the given sinusoidal wave is $y\left(x,t\right)=0.08\sin \left(2.5\pi x+6.0\pi t\right)$.

(b)

To determine

The expression for $y$ as a function of $x$ and $t$ for the given sinusoidal wave for given condition.

Answer to Problem 7P

The expression for $y$ as a function of $x$ and $t$ for the given sinusoidal wave for given condition is $y\left(x,t\right)=0.08\sin \left(2.5\pi x+6.0\pi t-0.25\pi \right)$.

Explanation of Solution

Given information:

The value of $A$ is $8.0\text{cm}$, the value of $\lambda$ is $80.0\text{cm}$, the value of $f$ is $3.0\text{Hz}$ and $y\left(x,0\right)$ is $0$ at $x=10.0\text{cm}$.

Rewrite the equation (II).

$y\left(x,t\right)=\left(0.08\text{m}\right)\sin \left(2.5\pi {\text{m}}^{-1}x+\left(6.0\pi {\text{s}}^{-1}\right)t+\varphi \right)$ (II)

Apply the condition $y\left(x,0\right)=0$ at $x=10.0\text{cm}$ in equation (II).

$\begin{array}{c}0=\left(0.08\text{m}\right)\sin \left(2.5\pi {\text{m}}^{-1}\times \left(10.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)+6.0\pi \times 0+\varphi \right)\\ 0=\sin \left(0.25\pi +\varphi \right)\\ 0.25\pi +\varphi =0\\ \varphi =-0.25\pi \end{array}$

Substitute $-0.25\pi$ for $\varphi$ in equation (II).

$\begin{array}{c}y\left(x,t\right)=\left(0.08\text{m}\right)\sin \left(2.5\pi {\text{m}}^{-1}x+\left(6.0\pi {\text{s}}^{-1}\right)t-0.25\pi \right)\\ =0.08\sin \left(2.5\pi x+6.0\pi t-0.25\pi \right)\end{array}$

Conclusion:

Therefore, the expression for $y$ as a function of $x$ and $t$ for the given sinusoidal wave for given condition is $y\left(x,t\right)=0.08\sin \left(2.5\pi x+6.0\pi t-0.25\pi \right)$.