Question

The equation of a transverse wave traveling along a string is

$\mathit{y}\mathbf{=}\left(2.0\mathrm{mm}\right)\mathbf{sin}\left[\left(20m^{-1}\right)x-\left(600s^{-1}\right)t\right]$

Find the (a) amplitude, (b) frequency, (c) velocity (including sign), and (d) wavelength of the wave. (e) Find the maximum transverse speed of a particle in the string.

Short answer

a) The amplitude of a wave traveling along a string is 2.0 mm.

b) The frequency of a wave traveling along a string is 90 Hz.

c) The velocity (including sign) of a wave traveling along a string is +30m/s.

d) The wavelength of a wave traveling along a string is 31 cm.

e) The maximum transverse speed of a particle in the string is 1.2 m/s.

Step-by-step solution

The given data

The wave equation is,$y=(2.0\mathrm{mm})\sin \left((20{\mathrm{m}}^{-1})\mathrm{x}-\left(600{\mathrm{s}}^{-1}\right)\mathrm{t}\right)$

Understanding the concept of the wave equation

We can find the values of amplitude, angular frequency, and wave constant by comparing the given equation with the general equation of the wave. Using them, we can find the frequency, wavelength, and velocity of the wave and maximum speed of the particle on the string using corresponding relations.

Formulae:

The wavenumber of the wave,

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

The angular frequency of the wave,

$\omega =2\mathrm{\pi f}$ (ii)

The transverse speed of the wave,

$v_m=\omega y_m$ (iii)

The speed of the wave,

$v=\lambda f$ (iv)

Here Here f is the frequency of the wave $\lambda$is the wavelength of the wave and $y_m$is the amplitude of the wave.

a) Calculation of amplitude

The general equation of a translational wave on a string is,

$y=y_m\sin \left(kx-\omega t\right)$ (1)

The given equation of the wave is,

$y=(2.0\mathrm{mm})\sin \left((20{\mathrm{m}}^{-1})\mathrm{x}-\left(600{\mathrm{s}}^{-1}\right)\mathrm{t}\right)$ (2)

Comparing equations (1) and (2), we can find the value of amplitude as:

$y_m=2.0\mathrm{mm}$

Therefore, the amplitude of a wave traveling along a string is 2.0 mm.

b) Calculation of frequency

Comparing equations (1) and (2), we can find the angular frequency as:

$\mathrm{\omega }=600{\mathrm{s}}^{-1}$

From equation (i) and the given values, the frequency of the wave can be calculated as,

$$\begin{gathered} \mathrm{f}=\frac{600}{2\mathrm{\pi }} \\ =95.45 \\ \cong 95\mathrm{Hz} \end{gathered}$$

Therefore, the frequency of a wave traveling along a string is 95 Hz.

c) Calculation of the velocity

Comparing equations (1) and (2), we can find the wavenumber as:

$k=20{\mathrm{m}}^{-1}$

Substitute the values in equation (ii), and the wavelength can be calculated as:

$$\begin{gathered} \lambda =\frac{2\mathrm{\pi }}{20} \\ =0.314m \end{gathered}$$

Substitute the values in equation (iii), and we can calculate the velocity of the wave as:

$$\begin{gathered} \mathrm{v}=95.54\left(0.314\right) \\ =30\mathrm{m}/\mathrm{s} \end{gathered}$$

Since the wave is traveling in the +x direction, velocity is in the +x direction.

Therefore, the velocity (including sign) of a wave traveling along a string is +30 m/s.

d) Calculation of wavelength

From part (c), the wavelength we calculated is:

$$\begin{gathered} \mathrm{\lambda }=0.314 \\ \cong 0.31\mathrm{m} \\ =31\mathrm{cm} \end{gathered}$$

Therefore, the wavelength of a wave traveling along a string is 31 cm.

e) Calculation of maximum transverse speed

Substitute the values in equation (iv), and the maximum speed of a wave can be calculated as:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{m}}=0.002\left(600\right) \\ =1.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the maximum transverse speed of a particle in the string is 1.2 m/s.