1. Wave speeds on the three strings

All three strings are connected in series, so the tension $T$ is the same in each string.

Wave speed on a string: $v=\sqrt{\frac{T}{\mu}}$

Thus,

• For $S_1$ with density $\mu$: $v_1=\sqrt{\frac{T}{\mu}}$
• For $S_2$ with density $4\mu$: $v_2=\sqrt{\frac{T}{4\mu}}=\frac{v_1}{2}$
• For $S_3$ with density $16\mu$: $v_3=\sqrt{\frac{T}{16\mu}}=\frac{v_1}{4}$

The incident wave on $S_1$ is $y=y_0\cos(\omega t-kx)$ So on $S_1$, $k=\frac{\omega}{v_1}$

Hence on other strings, $k_2=\frac{\omega}{v_2}=\frac{\omega}{v_1/2}=2k$ $k_3=\frac{\omega}{v_3}=\frac{\omega}{v_1/4}=4k$

2. Reflection and transmission at point $P$

At $P$, wave goes from lighter string $S_1$ to heavier string $S_2$.

For a wave incident from a lighter to a heavier medium, the reflected wave suffers a phase reversal of $\pi$.

Also, reflected wave must travel in the negative $x$-direction, so its form is $y\propto \cos(\omega t+kx+\phi)$ With phase reversal $\phi=\pi$, $y=\alpha_1 y_0\cos(\omega t+kx+\pi)$ where $\alpha_1>0$.

So A is correct.

Now for the transmitted wave at $P$:

• It travels in the positive $x$-direction.
• Frequency remains same.
• But wave number changes to $k_2=2k$ in string $S_2$.

So transmitted wave should be of the form $y=\alpha_2 y_0\cos(\omega t-2kx)$ not $y=\alpha_2 y_0\cos(\omega t-kx)$

Therefore B is incorrect.

3. Reflection and transmission at point $Q$

The wave reaching $Q$ is the transmitted wave in $S_2$, traveling toward $+x$. At $Q$, it goes from $S_2$ (density $4\mu$) to $S_3$ (density $16\mu$), i.e. again from lighter to heavier string.

So reflection at $Q$ also occurs with phase reversal of $\pi$.

But now the reflected wave is in string $S_2$, so it must travel in the negative $x$-direction and have wave number $k_2=2k$. Hence its form should be $y=\alpha_3 y_0\cos(\omega t+2kx+\pi)$ not $y=\alpha_3 y_0\cos(\omega t-kx+\pi)$

So C is incorrect.

4. Transmitted wave through $Q$

The wave transmitted through $Q$ enters string $S_3$.

• It travels in positive $x$-direction.
• Frequency remains $\omega$.
• Wave number becomes $k_3=4k$.

Hence the transmitted wave is $y=\alpha_4 y_0\cos(\omega t-4kx)$ with $\alpha_4>0$.

So D is correct.

5. Final answer

Correct options are: $\boxed{A,\ D}$

This matches the stored correct answer.