A transverse wave is represented by $y=A\sin(\omega t-kx)$. For what value of the wavelength is the wave veloc

Question

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?

Accepted Answer

1. **Determine Wave Velocity ($v$):** For a wave represented by $y = A\sin(\omega t - kx)$, the wave velocity is given by $v = \frac{\omega}{k}$ .
2. **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 $v_{\max} = A\omega$.
3. **Equate the Two Velocities:** The problem states that the wave velocity is equal to the maximum particle velocity:
   $$v = v_{\max}$$
   $$\frac{\omega}{k} = A\omega$$
4. **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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