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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?
A
A
B
$\frac{\pi A}{2}$
C
$\pi A$
D
$2\pi A$
Detailed Solution
Wave velocity: $v = \frac{\omega}{k}$
Particle velocity: $v_p = \frac{\partial y}{\partial t} = A\omega\cos(\omega t - kx)$; its maximum value is $A\omega$.
Setting them equal: $\frac{\omega}{k} = A\omega$
$\frac{1}{k} = A$; since $k = \frac{2\pi}{\lambda}$, $\frac{\lambda}{2\pi} = A$
$\lambda = 2\pi A$
Particle velocity: $v_p = \frac{\partial y}{\partial t} = A\omega\cos(\omega t - kx)$; its maximum value is $A\omega$.
Setting them equal: $\frac{\omega}{k} = A\omega$
$\frac{1}{k} = A$; since $k = \frac{2\pi}{\lambda}$, $\frac{\lambda}{2\pi} = A$
$\lambda = 2\pi A$