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NEET PHYSICSWAVEMedium

Question

A transverse wave is represented by y=Asin(ωtkx)y=A\sin(\omega t-kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

A

πA2\frac{\pi A}{2}

B

πA\pi A

C

2πA2\pi A

D

AA

Step-by-Step Solution

  1. Determine Wave Velocity (vv): For a wave represented by y=Asin(ωtkx)y = A\sin(\omega t - kx), the wave velocity is given by v=ωkv = \frac{\omega}{k} .
  2. Determine Maximum Particle Velocity (vmaxv_{\max}): The particle velocity vpv_p is the time derivative of the displacement yy: vp=yt=Aωcos(ωtkx)v_p = \frac{\partial y}{\partial t} = A\omega\cos(\omega t - kx) The maximum value of the particle velocity is the amplitude of this derivative, which is vmax=Aωv_{\max} = A\omega.
  3. Equate the Two Velocities: The problem states that the wave velocity is equal to the maximum particle velocity: v=vmaxv = v_{\max} ωk=Aω\frac{\omega}{k} = A\omega
  4. Solve for Wavelength (λ\lambda): Cancel ω\omega from both sides to get 1k=A\frac{1}{k} = A. We know that the wave number k=2πλk = \frac{2\pi}{\lambda} . Substitute this into the equation: 12πλ=A\frac{1}{\frac{2\pi}{\lambda}} = A λ2π=A\frac{\lambda}{2\pi} = A λ=2πA\lambda = 2\pi A

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.

PHYSICSWAVEtransverserepresentedyasinomegawavelengthvelocity

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A.$\sqrt{\nu}=\sqrt{\nu_1}+\sqrt{\nu_2}+\sqrt{\nu_3}$
B.$\nu=\nu_1+\nu_2+\nu_3$
C.$\frac{1}{\nu}=\frac{1}{\nu_1}+\frac{1}{\nu_2}+\frac{1}{\nu_3}$
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