Question

A $200\text{Hz}$ sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of $4\times {10}^{-3}\text{kg/m}$ and a tension of $40\text{N}$. At time $t=0$, the point $x=0$ has maximum displacement in the positive y-direction. Next when this point has zero displacement, the slope of the string is $\pi /20$. Which of the following expressions represents the displacement of string as a function of $x$ (in metres) and $t$ (in seconds)?

• A. $y=0.025\cos (400\pi t-2\pi x)$
• B. $y=0.0125\cos (400\pi t-4\pi x)$
• C. $y=0.025\cos (400\pi t-4\pi x)$
• D. $y=0.0125\cos (200\pi t-4\pi x)$

Answer 1

The correct option is B $y=0.0125\cos (400\pi t-4\pi x)$

Given data
Frequency $\left(f\right)=200\text{Hz}$
Linear mass density $\left(\mu \right)=4\times {10}^{-3}\text{kg/m}$
Tension $\left(T\right)=40\text{N}$
$\frac{dy}{dx}$ = slope of string $=\pi /20$ at $x=0$ when $y=0$
As we know,
$\omega =2\pi f=\left(2\pi \right)\left(200\right)=400\pi \text{rad/s}$
Wave velocity $\left(v\right)=\frac{\omega }{k}$
$\Rightarrow K=\frac{\omega }{v}=\frac{\omega }{\sqrt{T/\mu }}=\frac{400\pi }{\sqrt{40/4\times {10}^{-3}}}=\frac{400\pi }{100}=4\pi {\text{m}}^{-1}$
At zero displacement, velocity of particle $\left(v_p\right)=\omega A$
$\Rightarrow \omega A=-v\times \frac{dy}{dx}$
$\Rightarrow |A|=\frac{u\frac{dy}{dx}}{\omega }=\frac{\sqrt{\frac{T}{\mu }}}{\omega }\times \left(slope\right)$ where, $v$ is wave velocity.
$\Rightarrow |A|=\frac{\sqrt{{10}^4}}{400\pi }\times \frac{\pi }{20}=\frac{1}{80}=0.0125\text{m}$
Equation of a wave travelling in positive x-direction is given by
$y=A\cos (kx-\omega t)$ [since given, $y=A$ at $x=0,t=0$]
i.e $y=0.0125\cos (400\pi t-4\pi x)$
[since $\cos (-x)=\cos x$]
Option (b) is correct.