Question 7

MCQMEDIUM

Consider a system of three connected strings, $S_1$ , $S_2$ and $S_3$ with uniform linear mass densities $\mu$ kg/m, $4\mu$ kg/m and $16\mu$ kg/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 $O$ is connected to the free end of $S_1$ . The wave from the generator is represented by $y = y_0 \cos(\omega t - kx)$ cm, where $y_0, \omega$ and $k$ are constants of appropriate dimensions. Which of the following statements is/are correct:

(A)

When the wave reflects from $P$ for the first time, the reflected wave is represented by $y = \alpha_1 y_0 \cos(\omega t + kx + \pi)$ cm, where $\alpha_1$ is a positive constant.

(B)

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

(C)

When the wave reflects from $Q$ for the first time, the reflected wave is represented by $y = \alpha_3 y_0 \cos(\omega t - kx + \pi)$ cm, where $\alpha_3$ is a positive constant.

(D)

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

Detailed Solution

The speed of a wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$ . Since the strings are connected and in equilibrium, the tension $T$ is constant across all strings.

Determine wave speeds and wave numbers:

For $S_1$ : $v_1 = \sqrt{\frac{T}{\mu}}$ . Given wave is $y = y_0 \cos(\omega t - kx)$ , so $k_1 = k = \frac{\omega}{v_1}$ .
For $S_2$ : $v_2 = \sqrt{\frac{T}{4\mu}} = \frac{v_1}{2}$ . Thus, $k_2 = \frac{\omega}{v_2} = 2k$ .
For $S_3$ : $v_3 = \sqrt{\frac{T}{16\mu}} = \frac{v_1}{4}$ . Thus, $k_3 = \frac{\omega}{v_3} = 4k$ .

Analyze reflection at $P$ (Boundary between $S_1$ and $S_2$ ):

The wave travels from $S_1$ (rarer, higher speed) to $S_2$ (denser, lower speed).
Reflection coefficient $R = \frac{v_2 - v_1}{v_2 + v_1}$ . Since $v_2 < v_1$ , $R$ is negative. A negative coefficient implies a phase shift of $\pi$ .
The reflected wave travels in the $-x$ direction: $y_r = A_r \cos(\omega t + kx + \pi)$ . Thus, statement (A) is correct.

Analyze transmission through $P$ and $Q$ :

The wave transmitted through $P$ into $S_2$ has wave number $k_2 = 2k$ . Its form is $y_t = A_t \cos(\omega t - 2kx)$ . Statement (B) is incorrect because it uses $kx$ .
The wave incident on $Q$ from $S_2$ has the form $\sim \cos(\omega t - 2kx)$ . The wave transmitted into $S_3$ will have wave number $k_3 = 4k$ . Thus, $y_{tQ} = \alpha_4 y_0 \cos(\omega t - 4kx)$ . Statement (D) is correct.

Analyze reflection at $Q$ :

The reflected wave from $Q$ back into $S_2$ must travel in the $-x$ direction and have the wave number of $S_2$ , which is $2k$ . Statement (C) is incorrect because it uses $-kx$ (wrong direction) and $k$ (wrong wave number).

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