Question 53: Waves Optics

Question

Consider a system of three connected strings, $S_1, S_2$ and $S_3$ with uniform linear mass densities $\mu$ $\mathrm{kg} / \mathrm{m}, 4 \mu \mathrm{~kg} / \mathrm{m}$ and $16 \mu \mathrm{~kg} / \mathrm{m}$, respectively, as shown in the figure. $S_1$ and $S_2$ are connected at the point $P$, whereas $S_2$ and $S_3$ are connected at the point $Q$, and the other end of $S_3$ is connected to a wall. A wave generator 0 is connected to the free end of $S_1$. The wave from the generator is represented by $y=y_0 \cos (\omega t-k x) \mathrm{cm}$, where $y_0, \omega$ and $k$ are constants of appropriate dimensions. Which of the following statements is/are correct:

Accepted Answer

Given, wave eqn, $y=y_0 \cos (\omega t-k x) \mathrm{cm}$
$y_0, w, k$ are constants
we know, wave speed, $v=\sqrt{\frac{T}{\mu}}$, where $T=$ tension in string
$$ \begin{aligned} &	ext { So speeds : } v_1=\sqrt{\frac{T}{\mu}}, v_2=\sqrt{\frac{T}{4 \mu}}=\frac{1}{2} \sqrt{\frac{T}{\mu}}\ &\begin{aligned}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & \Rightarrow v_2=\frac{v_1}{2}, v_3=\sqrt{\frac{	au}{16 \mu}}=\frac{1}{4} \sqrt{\frac{	au}{\mu}} \ & \Rightarrow v_3=\frac{v_1}{4} \end{aligned} \end{aligned} $$
wave numbers: $k=\frac{\omega}{v}$
so $k_1=k, k_2=\frac{\omega}{v_2}=2 k$
and $k_3=\frac{\omega}{v_3}=4 k$
Reflection of wave in string from fixed end, reflected wave phase change by $\pi_0$.
And reflected wave from free (rarer) end no phase change.
$	herefore$ For medium $P Q\left(s_2ight)$
$$ k_{S_2}=k_{P Q}=2k $$
For medium, $S_3, k_{s_3}=4 k$
Option ( $A$ ) reflection from $P$, phase change by ' $\pi$, No Change in ' $K$ '. Hence, ( $A$ ) is the correct option.
In option (B), $k_{s_2}=2 k$ but in option given $k_s=k$. Hence, wrong option.
In option (C), the direction of the reflected wave is in the negative direction. Hence wrong,
option (D), no phase change in transmitted wave and given $k_{s_3}=4 k$
Here, correct option.
So (A), (D) are correct.

Answer

When the wave reflects from $P$ for the first time, the reflected wave is represented by $y=\alpha_1 \mathrm{y}_0 \cos (\omega t+k x+\pi) \mathrm{cm}$, where $\alpha_1$ is a positive constant.

Answer

When the wave transmits through $P$ for the first time, the transmitted wave is represented by $y=\alpha_2 y_0 \cos (\omega t-k x) \mathrm{cm}$, where $\alpha_2$ is a positive constant.

Answer

When the wave reflects from $Q$ for the first time, the reflected wave is represented by $y=\alpha_3 \mathrm{y}_0 \cos (\omega t-k x+\pi) \mathrm{cm}$, where $\alpha_3$ is a positive constant.

Answer

When the wave transmits through $Q$ for the first time, the transmitted wave is represented by $y=\alpha_4 y_0 \cos (\omega t-4 k x) \mathrm{cm}$, where $\alpha_4$ is a positive constant.

Waves Optics

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