Question

Two strings are attached to each other as shown in the figure. The linear mass density of the second string is $16$ times that of the first string and the boundary between the two strings is at $x=0$. The incident wave is given by $y=0.01\cos (20x-100t)$. Then, the ratio of amplitude of reflected wave to amplitude of transmitted wave is:
[Assume, tension in both the strings is same]

• A. $2:3$
• B. $3:2$
• C. $5:9$
• D. $9:5$

Answer 1

The correct option is B $3:2$

Given that, equation of incident wave
$y=0.01\cos (20x-100t)$
$=0.01\sin (20x-100t+\pi /2)$
Comparing with the general equation,
$y=A\sin (kx-\omega t+\varphi )$
we get,
$A=0.01\text{m}$ , $k=20{\text{m}}^{-1}$ , $\omega =100\text{rad/s}$
Let, the linear mass density of string $1$ be ${\mu }_1$ and that of string $2$ be $16{\mu }_1.$

Wave speed on $1^{st}$ string is given by $\left(v_1\right)=\sqrt{\frac{T}{{\mu }_1}}$
Wave speed on $2^{nd}$ string is given by $\left(v_2\right)=\sqrt{\frac{T}{16{\mu }_1}}$
$\Rightarrow v_2=\frac{v_1}{4}.......\left(1\right)$
Frequency of the wave depends on the vibrating source only
$\therefore \omega ={\omega }^{\prime }.....\left(2\right)$
Angular wave number of the wave on 1st string $\left(k\right)=\frac{\omega }{v_1}$
Angular wave number of the wave on 2nd string $\left(k^{\prime }\right)=\frac{\omega }{v_2}$
$\therefore k^{\prime }=4k$
Speed of incident wave on stretched string$1$ is
$v_1=\frac{\omega }{k}=\frac{100}{20}=5\text{m/s}$
From $\left(1\right)$, speed of wave in the stretched string$2$ is given by
$v_2=\frac{v_1}{4}=\frac{5}{4}\text{m/s}$

Now, Amplitude of the reflected wave

$A_r=|\frac{v_2-v_1}{v_1+v_2}|A_i$

Amplitude of transmitted wave

$A_t=\left(\frac{2v_2}{v_1+v_2}\right)A_i$

$\Rightarrow \frac{A_r}{A_t}=\left|\frac{v_2-v_1}{2v_2}\right|$

From the given data,

$\frac{A_r}{A_t}=\left|\frac{\frac{5}{4}-5}{\frac{5}{4}\times 2}\right|=\left|\frac{\frac{15}{4}}{\frac{10}{4}}\right|$
$\Rightarrow A_r:A_t=3:2$

Thus, option (b) is the correct answer.