Question

A traveling wave is produced on a long horizontal string by vibrating and ends up and down sinusoidally. The amplitude of vibration is $1.0cm$and the displacement becomes zero$200$ times per second. The linear mass density of the string is $0.10kg{m}^{-1}$ and it is kept under a tension of$90N$.

(a) Find the speed and the wavelength of the wave.

(b) Assume that the wave moves in the position 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$t=10ms.$

Answer 1

Step 1. Given data :

Amplitude,$A=1cm,$

Tension, $T=90N,$

Frequency,$f=200/2$

$=100Hz$

Mass density,$\mu =0.1kg/m$

Step 2. Find the speed $\left(v\right)$ and the wavelength$\left(\lambda \right)$ of the wave :

$v=\surd \left(\frac{T}{\mu }\right)$

$=\sqrt{\frac{90}{0.1}}$

$=30m/s$

Again,$\lambda =\frac{v}{f}$

$=\frac{30}{100}$

$=0.3m$

$=30cm$.

Hence, the speed of the wave is $30m/s$, and the wavelength is$30cm$.

Step 3. Find the wave equation :

Assume that the wave moves in the position x-direction and at $t=0$, the end $x=0$ is at its positive extreme position.

$y=A\cos (wt-kx)$

$\left[w=2\mathrm{\pi f},\mathrm{k}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\right]$

$y=(0.1)\cos (2\mathrm{\pi }(100)\mathrm{t}-\frac{2\mathrm{\pi }}{30}\mathrm{x})$

Hence, above is the required wave equation.

Step 3. Find the velocity and acceleration of the particle at$x=50$$cm$ at time$t=10ms.$

The velocity$\left(V\right)$of the particle,

$$\begin{gathered} y=A\cos (wt-kx) \\ V=\frac{dy}{dt} \\ =-A\sin (wt-kx)w \\ =-\left(1cm\right)\sin \left(2\mathrm{\pi }\right(100)\times 10\times {10}^{-3}-\frac{2\mathrm{\pi }}{30}\times 50)\frac{2\mathrm{\pi }}{10\times {10}^{-3}} \\ =-\left(1cm\right)\sin (2\mathrm{\pi }-\frac{10\mathrm{\pi }}{3})\frac{2\mathrm{\pi }}{10\times {10}^{-3}} \\ =-\left(1\mathrm{cm}\right)\sin (-\frac{4\mathrm{\pi }}{3})\frac{2\mathrm{\pi }}{10\times {10}^{-3}} \\ =-\left(1\mathrm{cm}\right)\sin (\mathrm{\pi }+\frac{\mathrm{\pi }}{3})\frac{2\mathrm{\pi }}{10\times {10}^{-3}} \\ =-\left(1\mathrm{cm}\right)\sin \left(\frac{\mathrm{\pi }}{3}\right)\frac{2\mathrm{\pi }}{10\times {10}^{-3}} \\ =-1\mathrm{cm}\times \frac{\sqrt{3}}{2}\frac{2\mathrm{\pi }}{10\times {10}^{-3}} \\ =-5.4\mathrm{m}/\mathrm{s} \end{gathered}$$

Now the acceleration $\left(a\right)$ is :

$$\begin{gathered} a=\frac{d^{2}y}{dt}=-A{w}^2\cos (wt-kx) \\ =-(1cm)\left(\frac{2\mathrm{\pi }}{10\times {10}^{-3}}\right)\cos (2\mathrm{\pi }\times 100\times 10\times {10}^{-3}-\frac{2\mathrm{\pi }}{30}\times 50) \\ \mathrm{a}=2\mathrm{km}/\mathrm{s} \end{gathered}$$

Hence, velocity of the particle is$-5.4m/s$ and acceleration is $2km/s$ .