Question

A $100Hz$ sinusoidal wave is travelling in the positive $x$ -direction along a string with a linear mass density of $3.5\times 10^{-3}kg/m$ and a tension of $35N$. 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 $\pi /20.$ The wave equation is

• A. $y=0.025\sin \left(200\pi t-2\pi x+\frac{\pi }{4}\right)$
• B. $y=0.025\sin \left(2\pi x-200\pi t+\frac{\pi }{2}\right)$
• C. $y=0.025\sin \left(2\pi x-200\pi t+\frac{\pi }{4}\right)$
• D. $y=0.025\sin \left(200\pi t-2\pi x+\frac{\pi }{2}\right)$

Answer 1

Answer: (D)

The speed of wave in string is
$V=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{35}{3.5\times 10^{-3}}}=100m/s$
$\lambda =\frac{v}{f}=\frac{100}{100}=1m$
Let us assume that
$y=A\sin (\omega t-kx+\theta )$
$t=0,x=0$ and $y=+A\therefore \theta =\frac{\pi }{2}$
Hence $y=A\sin \left(200\pi t-2\pi x+\frac{\pi }{2}\right)$
It is given that
$\frac{\partial y}{\partial x}=\frac{\pi }{20}$ at $t=\frac{T}{4}=\frac{1}{400}\sec$
${\frac{\partial y}{\partial x}]}_{t=\frac{1}{400}s}=A2\pi \cos \left(200\pi t-2\pi x+\frac{\pi }{2}\right)$
or $2A\pi =\frac{\pi }{20}$
$\Rightarrow A=\frac{1}{40}=0.025m$
Wave equation is
$y=0.025\sin \left(200\pi t-2\pi x+\frac{\pi }{2}\right)$