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(C) The maximum velocity of a particle in a wave is given by $v_{\max} = A \omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.G

Vedclass · Accepted Answer

$\frac{\lambda}{2 \pi}$

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(C) The maximum velocity of a particle in a wave is given by $v_{\max} = A \omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.Given that the maximum particle velocity is equal to the wave velocity $(v)$, we have $v_{\max} = v$.Substituting the expressions, we get $A \omega = v$.Since $\omega = 2 \pi f$ and the wave velocity $v = f \lambda$, we can write $\omega = 2 \pi (v / \lambda)$.Substituting this into the equation: $A 	imes (2 \pi v / \lambda) = v$.Dividing both sides by $v$, we get $A 	imes (2 \pi / \lambda) = 1$.Therefore, the amplitude $A = \frac{\lambda}{2 \pi}$.

When a sound wave of wavelength $\lambda$ is propagating in a medium, the maximum velocity of the particle is equal to the wave velocity. The amplitude of the wave is:

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