In a demonstration, a 1.2 kghorizontal rope is fixed in place at its two ends (x = 0 and x = 2.0m)and made to oscillate up and down in the fundamental mode, at frequency 5.0 Hz. At t = 0, the point at x = 1.0mhas zero displacement and is moving upward in the positive direction of a yaxis with a transverse velocity of 5.0m/s. What are (a) the amplitude of the motion of that point and (b) the tension in the rope? (c) Write the standing wave equation for the fundamental mode.

1. Mass of the rope is, m = 1.2 kg
2. Fundamental frequency on the rope is,$f_1=5.0\mathrm{Hz}$
3. At t = 0 , the point at x=1.0 m has zero displacement and velocity is,$v_m=5.0\mathrm{m}/\mathrm{s}$

A wave is an oscillation that moves through space while also transferring energy. Without moving the particles of the medium, the waves' motion results in an energy transfer. We can find the amplitude of the motion of the point on the rope from its velocity and frequency using the corresponding relation. Using the formula for the velocity of particles on the string we can find tension in the rope. From values of angular velocity, wave number, and amplitude we can write the standing wave equation for the fundamental mode.

Formula:

The velocity of an oscillation, $v=\frac{\omega }{k}$ …(i)

The maximum transverse speed, $v_m=\omega y_m$ …(ii)

The angular frequency of the wave,$\omega =2\pi f$ …(iii)

The speed of the wave, $v=f\lambda$ …(iv)

The velocity of the string, $v=\sqrt{\frac{T}{\mu }}$ …(v)

The wavenumber of the wave, $k=\frac{2\pi }{\lambda }$ …(vi)

From equation (iii), the angular velocity of the wave is given as:

$$\begin{gathered} \omega =2\pi \left(5\mathrm{Hz}\right) \\ =31.4\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, using value of angular frequency in equation (ii), we get the amplitude of the motion is given as:

$$\begin{gathered} y_m=\frac{5\mathrm{m}/\mathrm{s}}{31.4\mathrm{rad}/\mathrm{s}} \\ =0.159\mathrm{m} \\ ~0.16\mathrm{m} \end{gathered}$$

Therefore, the amplitude of the motion of the point on the rope is 0.16 m

Using equation (iv), the speed of waves on the rope is given as:

$v=f\left(2L\right)\left(\because \mathrm{\lambda }=\mathrm{first}\mathrm{fundamental}\mathrm{wavelength}=2\mathrm{L}\right)$

The tension of the rope using equation (v) is given as:

$$\begin{gathered} T=\frac{v^{2}m}{L}\left(\because \mu =\frac{m}{L}\right) \\ =\frac{{\left(2fL\right)}^{2}m}{L} \\ =4f^{2}Lm \\ =4{\left(5\mathrm{Hz}\right)}^2\left(2\mathrm{m}\right)(1.2\mathrm{kg}) \\ =240\mathrm{N} \\ =2.4\times {10}^2\mathrm{N} \end{gathered}$$

Therefore, tension in the rope is$2.4\times {10}^2\mathrm{N}$

The wave number using equation (vi) is given as:

$\begin{array}{l}k=\frac{2(3.14)}{2\left(2\mathrm{m}\right)}\left(\because \lambda =2\mathrm{L}\right)\\ =1.5{7\mathrm{m}}^{-1}\\ \end{array}$

Since, at t = 0 , the point at x = 1.0 m has zero displacement, this is a sine wave. Its equation is given as:

$$\begin{gathered} y=y_m\sin \left(kx\right)\sin \left(\omega t\right) \\ =(0.16\mathrm{m})sin\left[\left(1.57{\mathrm{m}}^{-1}\right)x\right]\sin \left(\left(31.4\frac{\mathrm{rad}}{\mathrm{s}}\right)t\right) \end{gathered}$$

Therefore, the standing wave equation for the fundamental mode is

$y=(0.16\mathrm{m})sin\left[\left(1.57{\mathrm{m}}^{-1}\right)x\right]\sin \left(\left(31.4\frac{\mathrm{rad}}{\mathrm{s}}\right)t\right)$