Question

A composition string is made up by joining two strings of different masses per unit length $⟶\mu$ and $4\mu .$ The composite string is under the same tension. A transverse wave pulse : $Y=\left(6mm\right)\sin (5t+40x)$, where '$t$' is in seconds and '$x$' in meters, is sent along the lighter string towards the joint. The joint is at $x=0.$ The equation of the wave pulse reflected from the joint is

• A. $\left(2mm\right)\sin (5t-40x)$
• B. $\left(4mm\right)\sin (40x-5t)$
• C. $-\left(2mm\right)\sin (5t-40x)$
• D. $\left(2mm\right)\sin (5t-10x)$

Answer 1

Answer: (C)

$-\left(2mm\right)\sin (5t-40x)$

Since the wave is travelling from low to more dense medium, thus there will be inversion of the reflected wave . Since it travels back from origin, thus $40x$ will become $-40x.$ Also, amplitude of reflected wave is given by:
$A_r=\frac{Z_1-Z_2}{Z_1+Z_2}A.Z=\mu c=\mu \sqrt{\frac{T}{\mu }}=\sqrt{\mu T}Thus{A}_r=\frac{\sqrt{\mu T}-\sqrt{4\mu T}}{\sqrt{\mu T}+\sqrt{4\mu T}}A=\frac{-1}{3}A$
Thus it become $-6/3=-2$.