Question

A transverse progressive wave is established on a $100m$ long rope of mass 4 kg under tension of $100N$. If $25$ wave pulses can accomodate on the string and maximum particle velocity of the string is 5cm/s and at $t=0$, particle at $x=0$ is at its mean position and going down find out the transverse wave equation. Assume sinusoidal wave going towards right.

• A. $y=\left(\frac{\pi }{100}\right)\sin \left(12.5\pi t-\frac{\pi x}{2}+\pi \right)$
• B. $y=\left(\frac{1}{500\pi }\right)\sin \left(25\pi t-\frac{\pi x}{2}+\pi \right)$
• C. $y=\left(\frac{1}{50\pi }\right)\sin \left(25\pi t-\frac{\pi x}{2}\right)$
• D. $y=\left(\frac{\pi }{10}\right)\sin \left(25\pi t-\frac{\pi x}{2}+\pi \right)$

Answer 1

Answer: (B)

$\lambda =\frac{100}{25}=4m$
$V=\sqrt{\frac{T}{u}}=\sqrt{\frac{100}{4/100}}$
$=\frac{100}{2}=50m/s$
$V=\frac{\omega }{K}$
$\Rightarrow 50=\frac{\omega }{2\pi /X}$
$\left\{\omega =50\times \frac{2\pi }{4}=25\pi \right\}$
Since at $t=0$ and at $x=0$ particle is going down. So $\varphi =\pi$
$\therefore$ Equation becomes
$y=A\sin (\omega t-Kx+\pi )$
$V_{\max }=A\omega$
$\frac{5}{100}=A\times 25\pi$
$A=\frac{1}{500\pi }m$
$\left\{y=\frac{1}{500\pi }\sin \left(25\pi t-\frac{\pi }{2}x+\pi \right)\right\}$