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NEET PHYSICSWAVEMedium

Question

The equation of a simple harmonic wave is given by y=3sinπ2(50tx)y=3\sin\frac{\pi}{2}(50t-x) where xx and yy are in meters and tt is in seconds. The ratio of maximum particle velocity to the wave velocity is:

A

2π2\pi

B

32π\frac{3}{2}\pi

C

3π3\pi

D

23π\frac{2}{3}\pi

Step-by-Step Solution

  1. Identify the Given Wave Equation: The equation is y=3sinπ2(50tx)y = 3\sin\frac{\pi}{2}(50t - x), which can be rewritten as y=3sin(25πtπ2x)y = 3\sin(25\pi t - \frac{\pi}{2}x).
  2. Compare with the Standard Wave Equation: The standard progressive wave equation is y=Asin(ωtkx)y = A\sin(\omega t - kx) . By comparing, we get:
  • Amplitude, A=3 mA = 3 \text{ m}
  • Angular frequency, ω=25π rad/s\omega = 25\pi \text{ rad/s}
  • Wave number, k=π2 m1k = \frac{\pi}{2} \text{ m}^{-1}
  1. Calculate Maximum Particle Velocity (vmaxv_{\max}): The maximum particle velocity is given by vmax=Aω=3×25π=75π m/sv_{\max} = A\omega = 3 \times 25\pi = 75\pi \text{ m/s}.
  2. Calculate Wave Velocity (vv): The wave velocity is given by v=ωk=25ππ/2=50 m/sv = \frac{\omega}{k} = \frac{25\pi}{\pi/2} = 50 \text{ m/s} .
  3. Find the Ratio: The ratio of maximum particle velocity to the wave velocity is: vmaxv=75π50=32π\frac{v_{\max}}{v} = \frac{75\pi}{50} = \frac{3}{2}\pi (Alternatively, this ratio can be directly found as Ak=3×π2=32πAk = 3 \times \frac{\pi}{2} = \frac{3}{2}\pi).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVEequationsimpleharmonicysinfracpitxmeters

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