Question

A traveling wave is produced on a long horizontal string by vibrating 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 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 $x=50cm$ at time $t=10ms$.

Answer 1

Step 1: Given Data:

Amplitude $A=0.01m$

Tension $T=90N$

Frequency $f=\frac{200}{2}$ (because in a single oscillation the displacement becomes zero twice)

$=100Hz$

Mass density $\rho =0.1kg{m}^{-1}$

Step 2: Finding speed and wavelength:

Assume the velocity as $v$ and wavelength as $\lambda$ then,

We know, $v=\sqrt{\frac{T}{\rho }}$

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

$\Rightarrow$ $v=30m{s}^{-1}$

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

$\Rightarrow$ $\lambda =\frac{30}{100}$

$\Rightarrow$ $\lambda =0.3m$

$\Rightarrow$ $\lambda =30cm$

Step 3: Writing the wave equation

General equation of wave:

$y(x,t)=A\sin (kx-\omega \varphi )$

For the given case, $x=0$, displacement is maximum.

As we know, $k=\frac{2\mathrm{\pi }}{\lambda }$ and $\omega =2\pi \nu$.

Thus, the wave equation takes the form

$y=\left(1cm\right)\cos 2\pi \{\left(\frac{t}{0.01s}\right)-\left(\frac{x}{30cm}\right)\}$

Step 4: Finding equations for velocity and acceleration of the particle:

Velocity is given by

$v=\frac{dy}{dt}$

$=\left(\frac{2\pi }{0.01}\right)\sin 2\pi \{\left(\frac{x}{30}\right)-\left(\frac{t}{0.01}\right)\}$

and, $a=\frac{dv}{dt}$

$=\left\{\frac{4{\pi }^2}{{\left(0.01\right)}^2}\right\}\cos 2\pi \left\{\left(\frac{x}{30}\right)-\left(\frac{t}{0.01}\right)\right\}$

Step 5: Calculating the values of velocity and acceleration:

For the values $x=50cm$

and time $t=0.01s$

$v=\left(\frac{2\pi }{0.01}\right)\sin 2\pi \{\left(\frac{50}{30}\right)-\left(\frac{0.01}{0.01}\right)\}$

$=\left(\frac{2\pi }{0.01}\right)\sin 2\pi \left\{\left(\frac{2}{3}\right)\right\}$

$=\left(\frac{2\pi }{0.01}\right)\sin \left(\frac{4\pi }{3}\right)$

$=200\pi \sin \frac{\pi }{3}$

$=200\pi \times \frac{\sqrt{3}}{2}$

$=5.4m{s}^{-1}$

Now, for acceleration

$a=\left\{\frac{4{\pi }^2}{{\left(0.01\right)}^2}\right\}\cos 2\pi \left\{\left(\frac{x}{30}\right)-\left(\frac{t}{0.01}\right)\right\}$

$=\left\{\frac{4{\pi }^2}{{\left(0.01\right)}^2}\right\}\cos 2\pi \left\{\left(\frac{50}{30}\right)-\left(\frac{0.01}{0.01}\right)\right\}$

$=4{\pi }^2\times {10}^4\times 0.5$

$=2\times {10}^{3}m{s}^{-2}$

Hence, the velocity and acceleration are $\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$ and $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{2}}$.