JEE Advanced 2025 Paper 1 Q07 Physics Waves & Optics Waves on String Hard

JEE Advanced 2025 Paper 1 · Q07 · Waves on String

Three strings $S_1$, $S_2$, $S_3$ with linear mass densities $\mu$, $4\mu$, and $16\mu$ kg/m respectively are connected end-to-end. $S_1$ and $S_2$ join at point $P$; $S_2$ and $S_3$ join at $Q$. The far end of $S_3$ is fixed to a wall. A wave generator at the free end of $S_1$ produces $y = y_0\cos(\omega t - k x)$ cm (incident, amplitude $y_0$). All strings have the same tension. Which of the following are correct?

  1. A. Reflected wave at $P$ (first time): $y = \alpha_1 y_0 \cos(\omega t + k x + \pi)$ cm with $\alpha_1 > 0$.
  2. B. Transmitted wave through $P$ (first time): $y = \alpha_2 y_0 \cos(\omega t - k x)$ cm with $\alpha_2 > 0$.
  3. C. Reflected wave at $Q$ (first time): $y = \alpha_3 y_0 \cos(\omega t - k x + \pi)$ cm with $\alpha_3 > 0$.
  4. D. Transmitted wave through $Q$ (first time): $y = \alpha_4 y_0 \cos(\omega t - 4 k x)$ cm with $\alpha_4 > 0$.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, D

Solution

Wave speed $v \propto 1/\sqrt{\mu}$. For $S_1$ ($\mu$), $S_2$ ($4\mu$), $S_3$ ($16\mu$), $v_2 = v_1/2$ and $v_3 = v_1/4$, while frequency is preserved across boundaries. So at boundary $P$ ($S_1 \to S_2$), the reflection coefficient $r = (v_2 - v_1)/(v_2 + v_1) = (v_1/2 - v_1)/(v_1/2 + v_1) = -1/3$. Negative $r$ means the reflected wave is inverted, i.e., a phase shift of $\pi$. (A) writes the reflected wave with $\cos(\omega t + k x + \pi)$, correctly capturing the $\pi$ phase flip; $\alpha_1 = 1/3 > 0$. (A) TRUE. (B) Transmitted wave into $S_2$ has wavenumber $k_2 = \omega/v_2 = 2k$ (since $v$ halved, $k$ doubles). So the transmitted form should be $y = t_1 y_0 \cos(\omega t - 2 k x)$, not $\cos(\omega t - k x)$. (B) FALSE. (C) At $Q$ ($S_2 \to S_3$): $r = (v_3 - v_2)/(v_3 + v_2) = (v_1/4 - v_1/2)/(v_1/4 + v_1/2) = -1/3$ (negative), so reflected wave is also inverted. The wave traveling back in $S_2$ has wavenumber $2k$. But the option writes $\cos(\omega t - k x + \pi)$, keeping the original $k$ and the same sign of $x$ — incorrect both in wavenumber and direction. (C) FALSE. (D) Transmitted into $S_3$: wavenumber $k_3 = \omega/v_3 = 4k$, same direction ($-x$ propaga…[truncated]

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