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 $\frac{\pi }{20}$. Which of the following expressions respresent the displacement of string as a function of $x$ (in metre) and $t$ (in seconds).

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

Answer 1

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

Given data :
frequency $\left(f\right)=200\text{Hz}$
linear mass density, $\left(\mu \right)=4\times {10}^{-3}\text{kg/m}$
Tension $\left(T\right)=40\text{N}$
slope of string when $y=0$ is given by
$\frac{dy}{dx}=\frac{\pi }{20}$
As we know
$\omega =2\pi f=\left(2\right(\pi \left)\right(200)=400\pi \text{rad/s}$
Wave velocity $\left(v\right)=\frac{\omega }{k}$
$\Rightarrow k=\frac{\omega }{v}=\frac{\omega }{\sqrt{\frac{T}{\mu }}}=\frac{400\pi }{\sqrt{\frac{40}{4\times {10}^{-3}}}}=\frac{400\pi }{100}=4\pi {\text{m}}^{-1}$

At zero displacement
velocity of particle $\left(v_p\right)=-\omega A$
$\Rightarrow -\omega A=-(v\frac{dy}{dx})$
$\Rightarrow A=\frac{v\frac{dy}{dx}}{\omega }=\frac{\left(\frac{dy}{dx}\right)}{k}$ where $v=$ wave velocity
$\Rightarrow A=\frac{1}{4\pi }\times \frac{\pi }{20}=\frac{1}{80}=0.0125\text{m}$
So,
Equation of wave becomes,
$y=0.0125\cos (4\pi x-400\pi t)$