Question

A transverse progressive wave is established on a $100\, m$ long rope of mass 4 kg under tension of $100\, N$. 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)

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