Question

A $100\, Hz$ 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 $35\, N$. 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 $\pi / 20 .$ 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
$V=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{35}{3.5 \times 10^{-3}}}=100\, m / s$
$\lambda=\frac{v}{f}=\frac{100}{100}=1\, m$
Let us assume that
$y=A \sin (\omega t-k x+\theta)$
$t=0, x=0$ 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 $t=\frac{T}{4}=\frac{1}{400} \sec$
$\left.\frac{\partial y}{\partial x}\right]_{t=\frac{1}{400} s}=A 2 \pi \cos \left(200 \pi t-2 \pi x+\frac{\pi}{2}\right)$
or $2 A \pi=\frac{\pi}{20}$
$\Rightarrow A=\frac{1}{40}=0.025\, m$
Wave equation is
$y=0.025 \sin \left(200 \pi t-2 \pi x+\frac{\pi}{2}\right)$