Question

A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is 1⋅0 and the displacement becomes zero 200 times per second. The linear mass density of the string is 0⋅10 kg m⁻¹ and it is kept under a tension of 90 N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10 ms.

Answer 1

Given,
Amplitude of the wave = 1 cm
Frequency of the wave, $f=\frac{200}{2}=100\mathrm{Hz}$
Mass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N

(a) Velocity of the wave is given by
$v=\sqrt{\frac{T}{m}}$
Thus, we have:
$$\begin{gathered} v=\sqrt{\left(\frac{90}{0.1}\right)}=30\mathrm{m}/\mathrm{s} \\ \mathrm{Now}, \\ \mathrm{Wavelength},\lambda =\frac{v}{f}=\frac{30}{100}=0.3\mathrm{m} \\ \Rightarrow \mathrm{\lambda }=30\mathrm{cm} \end{gathered}$$

(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by
$y=\left(1\mathrm{cm}\right)\cos 2\pi \left\{\left(\frac{t}{0.01\mathrm{s}}\right)-\left(\frac{x}{30\mathrm{cm}}\right)\right\}$ ...(1)

(c) Using $\cos \left(-\theta \right)=\cos \theta$ in equation (1), we get:
$y=1\cos 2\pi \left(\frac{x}{30}-\frac{t}{0.01}\right)$
$$\begin{gathered} \mathrm{Velocity},v=\frac{dy}{dt} \\ \Rightarrow v=\left(\frac{2\pi }{0.01}\right)\sin 2\pi \left\{\frac{x}{30}-\frac{t}{0.01}\right\} \\ \mathrm{And}, \\ \mathrm{Acceleration},a=\frac{d\nu }{dt} \\ \Rightarrow a=\left\{\frac{4{\pi }^2}{{\left(0.01\right)}^2}\right\}\cos 2\pi \left\{\left(\frac{x}{30}\right)-\left(\frac{t}{0.01}\right)\right\} \\ \mathrm{When}x=50\mathrm{cm},t=10\mathrm{ms}=10\times {10}^{-3}\mathrm{s}. \end{gathered}$$

Now,

$$\begin{gathered} v=\left(\frac{2\pi }{0.01}\right)\sin 2\pi \left\{\left(\frac{5}{3}\right)-\left(\frac{0.01}{0.01}\right)\right\} \\ =\left(\frac{2\pi }{0.01}\right)\sin \left(2\pi \times \frac{2}{3}\right) \\ =-\left(\frac{2\mathrm{\pi }}{0.01}\right)\sin \frac{4\pi }{3} \\ =-200\pi \sin \frac{\pi }{3} \\ =-200\pi \times \frac{\sqrt{3}}{2} \\ =-544\mathrm{cm}/\mathrm{s} \\ =-5.4\mathrm{m}/\mathrm{s} \end{gathered}$$
In magnitude, v = 5.4 m/s.
Similarly,
$$\begin{gathered} a=\left\{\frac{4{\pi }^2}{{\left(0.01\right)}^2}\right\}\cos 2\pi \left\{\left(\frac{5}{3}\right)-1\right\} \\ =4{\pi }^2\times {10}^4\times \frac{1}{2} \\ \approx 2\times {10}^5\mathrm{cm}/{\mathrm{s}}^2\mathrm{or}2\mathrm{km}/{\mathrm{s}}^2 \end{gathered}$$