Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

$\frac{\pi A}{2}$

$\pi A$

$2\pi A$

$A$

Step-by-Step Solution

Determine Wave Velocity ( $v$ ): For a wave represented by $y = A\sin(\omega t - kx)$ , the wave velocity is given by $v = \frac{\omega}{k}$ .
Determine Maximum Particle Velocity ( $v_{\max}$ ): The particle velocity $v_p$ is the time derivative of the displacement $y$ : $v_p = \frac{\partial y}{\partial t} = A\omega\cos(\omega t - kx)$ The maximum value of the particle velocity is the amplitude of this derivative, which is $v_{\max} = A\omega$ .
Equate the Two Velocities: The problem states that the wave velocity is equal to the maximum particle velocity: $v = v_{\max}$ $\frac{\omega}{k} = A\omega$
Solve for Wavelength ( $\lambda$ ): Cancel $\omega$ from both sides to get $\frac{1}{k} = A$ . We know that the wave number $k = \frac{2\pi}{\lambda}$ . Substitute this into the equation: $\frac{1}{\frac{2\pi}{\lambda}} = A$ $\frac{\lambda}{2\pi} = A$ $\lambda = 2\pi A$

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