Chapter 11, Problem 25P

(a)

To determine

The amplitude of the transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$.

Answer to Problem 25P

The amplitude of the transverse wave on the string is $0.35\text{mm}$.

Explanation of Solution

Write the standard expression of a transverse wave travelling along $x$ direction.

$y\left(x,t\right)=A\sin \left(kx-\omega t\right)$ (I)

Here, $A$ is the amplitude of wave, $k$ is the wave vector, $\omega$ is the angular velocity, $x$ is position of wave at any time $t$ and $y\left(x,t\right)$ is the displacement of wave at position $x$ and time $t$.

Write the equation of wave given in question.

$y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ (II)

Conclusion:

Compare equation (II) with (I) to get $A$.

$A=0.35\text{mm}$

Therefore, the amplitude of the transverse wave on the string is $0.35\text{mm}$.

(b)

To determine

The wavelength of the transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$.

Answer to Problem 25P

The wavelength of the transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ is $6.0\text{m}$.

Explanation of Solution

Write the standard expression of a transverse wave travelling along $x$ direction.

$y\left(x,t\right)=A\sin \left(kx-\omega t\right)$ (I)

Here, $A$ is the amplitude of wave, $k$ is the wave vector, $\omega$ is the angular velocity, $x$ is position of wave at any time $t$ and $y\left(x,t\right)$ is the displacement of wave at position $x$ and time $t$.

Write the equation of wave given in question.

$y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ (II)

Expand above equation to compare with standard form given in equation (I).

$y\left(x,t\right)=\left(0.35\text{mm}\right)\left[\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)x-\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left(66\frac{\text{m}}{\text{s}}\right)t\right]$ (III)

Write the expression for wavelength of wave.

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

Conclusion:

Compare equation (III) with (I) to get $k$.

$k=\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}$

Substitute $\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}$ for $k$ in equation (IV) to get $\lambda$.

$\begin{array}{c}\lambda =\frac{2\pi }{\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}}\\ =6.0\text{m}\end{array}$

Therefore, the wavelength of the transverse wave on a string is $6.0\text{m}$.

(c)

To determine

The frequency of the transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$.

Answer to Problem 25P

The frequency of the transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ is $11\text{Hz}$.

Explanation of Solution

Write the standard expression of a transverse wave travelling along $x$ direction.

$y\left(x,t\right)=A\sin \left(kx-\omega t\right)$ (I)

Here, $A$ is the amplitude of wave, $k$ is the wave vector, $\omega$ is the angular velocity, $x$ is position of wave at any time $t$ and $y\left(x,t\right)$ is the displacement of wave at position $x$ and time $t$.

Write the equation of wave given in question.

$y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ (II)

Expand above equation to compare with standard form given in equation (I).

$y\left(x,t\right)=\left(0.35\text{mm}\right)\left[\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)x-\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left(66\frac{\text{m}}{\text{s}}\right)t\right]$ (III)

Write the expression for frequency of wave.

$f=\frac{\omega }{2\pi }$ (V)

Conclusion:

Compare equation (III) with (I) to get $\omega$.

$\omega =\left(\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)66\frac{\text{m}}{\text{s}}\right)$

Substitute $\left(\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)66\frac{\text{m}}{\text{s}}\right)$ for $\omega$ in equation (V) to get $f$.

$\begin{array}{c}f=\frac{\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\times 66\frac{\text{m}}{\text{s}}}{2\pi }\\ =11\text{Hz}\end{array}$

Therefore, the frequency of the transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ is $11\text{Hz}$.

(d)

To determine

The direction of the transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$.

Answer to Problem 25P

The transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ travels in the $+x$ direction.

Explanation of Solution

Write the standard expression of a transverse wave travelling along $x$ direction.

$y\left(x,t\right)=A\sin \left(kx-\omega t\right)$ (I)

Here, $A$ is the amplitude of wave, $k$ is the wave vector, $\omega$ is the angular velocity, $x$ is position of wave at any time $t$ and $y\left(x,t\right)$ is the displacement of wave at position $x$ and time $t$.

Write the equation of wave given in question.

$y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ (II)

Equation (I) represents a transverse traveling along $+x$ direction. This is possible if sign of term $\omega t$ is negative and sign of $kx$ is positive.

Conclusion:

The wave in equation (II) represents same type of wave that travels in $+x$ direction.

Therefore, the transverse wave on a string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ travels in the $+x$ direction.

(e)

To determine

The maximum transverse speed of a point on the string.

Answer to Problem 25P

The maximum transverse speed of a point on the. string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ is $24\text{mm}/\text{s}$.

Explanation of Solution

Write the standard expression of a transverse wave travelling along $x$ direction.

$y\left(x,t\right)=A\sin \left(kx-\omega t\right)$ (I)

Here, $A$ is the amplitude of wave, $k$ is the wave vector, $\omega$ is the angular velocity, $x$ is position of wave at any time $t$ and $y\left(x,t\right)$ is the displacement of wave at position $x$ and time $t$.

Write the equation of wave given in question.

$y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ (II)

Expand above equation to compare with standard form given in equation (I).

$y\left(x,t\right)=\left(0.35\text{mm}\right)\left[\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)x-\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left(66\frac{\text{m}}{\text{s}}\right)t\right]$ (III)

Write the expression for maximum transverse speed of a point on the string.

$v_{\text{max}}=\omega A$ (VI)

Conclusion:

Substitute $\frac{\pi }{3.0\text{m}}\left(\frac{66\text{m}}{\text{s}}\right)$ for $\omega$ and $0.35\text{mm}$ for $A$ in above equation to get $v_{\text{max}}$.

$\begin{array}{c}{v}_{\text{max}}=\left(\frac{\pi }{3.0\text{m}}\left(\frac{66\text{m}}{\text{s}}\right)\right)\left(0.35\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\\ =24\text{mm}/\text{s}\end{array}$

Therefore, the maximum transverse speed of a point on the. string described by $y\left(x,t\right)=\left(0.35\text{mm}\right)\left\{\left(\frac{\pi }{3.0}\frac{\text{rad}}{\text{m}}\right)\left[x-\left(66\frac{\text{m}}{\text{s}}\right)\right]t\right\}$ is $24\text{mm}/\text{s}$.