Question

A sinusoidal transverse wave of amplitude${\mathit{y}}_{\mathbf{m}}$ and wavelength travels on a stretched cord. (a) Find the ratio of the maximum particle speed (the speed with which a single particle in the cord moves transverse to the wave) to the wave speed. (b) Does this ratio depend on the material of which the cord is made?

Short answer

a) The ratio of the maximum particle speed to the wave speed is$2{\mathrm{\pi y}}_{\mathrm{m}}/\mathrm{\lambda }$

b) This ratio does not depend on the material of which the cord is made

Step-by-step solution

Given data

Amplitude of a wave is$y_m$

Wavelength of the wave is $\lambda$

Understanding the concept of the wave equation

We use the equation of displacement of the wave and wave speed to take their ratio. From this, we can find the ratio of maximum particle speed to wave speed and find whether this ratio depends on the material of which the cord is made or not.

Formula:

The general expression of the wave,$y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)..........\left(1\right)$

The velocity of a wave, $v=\lambda /T........\left(2\right)$

The wavelength of a wave, $\lambda =2\mathrm{\pi }/\mathrm{k}.....\left(3\right)$

The time period of an oscillation, $T=2\mathrm{\pi }/\mathrm{\omega }.........\left(4\right)$

Step 3(a): Calculation of ratio of maximum particle speed to wave speed

Using equation (1), the velocity of the wave is given as the derivative of the displacement as:

$$\begin{gathered} u\left(x,t\right)=\frac{dy}{dt} \\ =\frac{d\left[y_m\sin \left(kx-\omega t\right)\right]}{dt} \\ =-\omega y_m\cos \left(kx-\omega t\right) \end{gathered}$$

And its maximum value occurs when cosine value term is 1; hence the maximum particle speed is given as;

$u_m=\omega y_m$

Again, substituting values of equations (3) and (4) in equation (2), we get the speed of the wave as:

$$\begin{gathered} v=\frac{{\displaystyle \frac{2\mathrm{\pi }}{k}}}{\frac{2\mathrm{\pi }}{\omega }} \\ =\frac{\omega }{k} \end{gathered}$$

Thus, the required ratio is given as;

$$\begin{gathered} \frac{u_m}{v}=\frac{\omega y_m}{{\displaystyle \frac{\omega }{k}}} \\ =k{y}_m \\ =\frac{2{\mathrm{\pi y}}_{\mathrm{m}}}{\lambda }(fromequation\left(3\right)) \end{gathered}$$

Therefore, the ratio of the maximum particle speed to the wave speed is$\frac{2{\mathrm{\pi y}}_{\mathrm{m}}}{\lambda }$

Step 4(b): Checking whether the ration of speeds depend on the material

The ratio of the maximum particle speed to the wave speed is given as:

$\frac{u_m}{v}=\frac{2{\mathrm{\pi y}}_{\mathrm{m}}}{\lambda }$

This ratio depends only on amplitude to the wavelength.

Therefore, the ratio is not dependent on the material of which the cord is made.