Question

A harmonic wave is travelling on string 1. At a junction with string 2, it is partly reflected and partly transmitted. The linear mass density of the second string is four times that of the first string and the boundary between the two strings is at x = 0. If the expression for the incident wave is $y_1=A_1\cos (k_{1}x-{\omega }_{1}t)$

What is the equation for the reflected wave in terms of $A_1,k_1$ and ${\omega }_1$ ?

• A. $y=\frac{A_1}{4}\cos (k_{1}x+{\omega }_{1}t+\pi )$
• B. $y=\frac{A_1}{6}\cos (k_{1}x+{\omega }_{1}t+\pi )$
• C. $y=\frac{A_1}{3}\cos (k_{1}x+{\omega }_{1}t+\pi )$
• D. $y=\frac{A_1}{3}\cos (2k_{1}x+{\omega }_{1}t+\pi )$

Answer 1

Answer: (C)

$y=\frac{A_1}{3}\cos (k_{1}x+{\omega }_{1}t+\pi )$

Since $v=\sqrt{T/\mu }$ $T_2=T_{1}and{\mu }_2=4{\mu }_1$
we have $v_2=\frac{v_1}{2}$ ...........(i)
The frequency does not change that is
${\omega }_1={\omega }_2$ ................(ii)
Also because $k=\omega /v$ the wave numbers of the harmonic waves in the two strings are related by
$k_2=\frac{{\omega }_2}{v_2}=\frac{{\omega }_1}{v_1/2}=2\frac{{\omega }_1}{v_1}=2k_1$ ...................(iii)
The amplitudes are
$A_1=\left(\frac{2v_2}{v_1+v_2}\right)A_1=\left[\frac{2\left(v_1/2\right)}{v_1+\left(v_1/2\right)}\right]A_1=\frac{2}{3}{A}_1$ .............(iv)
and $A_1=\left(\frac{v_2-v_1}{v_1+v_2}\right)A_1=\left[\frac{\left(v_1/2\right)-v_1}{v_1+\left(v_1/2\right)}\right]A_1=\frac{A_1}{3}$ .................(v)
Now with equation (ii), (iii) and (iv) the transmitted wave can be written as
$y_1=\frac{2}{3}{A}_1\cos (2k_{1}x-{\omega }_{1}t)$ Ans
Similarly the reflected wave can be expressed as
$=\frac{A_1}{3}\cos (k_{1}x+{\omega }_{1}t+\pi )$