Question

Two strings are attached to each other as shown in the figure. 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 $x=0$. The wave equation of the incident wave at the boundary is $y_i=A_i\cos (k_{1}x-{\omega }_{1}t)$. Then, which of the following correctly represents the equation of the transmitted wave? Assume, tension in the two strings are same.

• A. $y_t=\frac{3}{2}{A}_i\cos (k_{1}x-{\omega }_{1}t)$
• B. $y_t=\frac{2}{3}{A}_i\cos (k_{1}x-{\omega }_{1}t)$
• C. $y_t=\frac{3}{2}{A}_i\cos (2k_{1}x-{\omega }_{1}t)$
• D. $y_t=\frac{2}{3}{A}_i\cos (2k_{1}x-{\omega }_{1}t)$

Answer 1

The correct option is D $y_t=\frac{2}{3}{A}_i\cos (2k_{1}x-{\omega }_{1}t)$

Given that ,
$y_i=A_i\cos (k_{1}x-{\omega }_{1}t)$ is the equation of incident wave.

Let the linear mass density of the 1st string be $\mu$ and of the 2nd string be $4\mu$.

Wave speed in 1st stretched string $\left(v_1\right)=\sqrt{\frac{T}{\mu }}$

Wave speed in 2nd stretched string $\left(v_2\right)=\sqrt{\frac{T}{4\mu }}$
Tension in the two strings are same
$\Rightarrow v_2=\frac{v_1}{2}......\left(1\right)$
The frequency of the wave depends on the vibrating source only.
So, ${\omega }_2={\omega }_1......\left(2\right)$

Angular wave number on 1st string $\left(k_1\right)=\frac{{\omega }_1}{v_1}$
Angular wave number on 2nd string $\left(k_2\right)=\frac{{\omega }_2}{v_2}$

Using $\left(1\right)$ in the above formula we get,
$k_2=2k_1......\left(3\right)$

Amplitude of the transmitted wave is given by
$A_t=\left(\frac{2v_2}{v_1+v_2}\right)A_i$

From the data obtained, we can rewrite the above equation as
$A_t=\left(\frac{2\left(\frac{v_1}{2}\right)}{v_1+\left(\frac{v_1}{2}\right)}\right)A_i$
$\Rightarrow A_t=\frac{2}{3}{A}_i...........\left(4\right)$
Now from equations $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$ equation of the transmitted wave can be written as

$y_t=\frac{2}{3}{A}_i\cos (2k_{1}x-{\omega }_{1}t)$

Hence option (d) is the correct answer.