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(C) The maximum particle velocity $v_{\max}$ for a wave is given by $v_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequ

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The maximum particle velocity $v_{\max}$ for 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 $v = f \lambda$,we can write $\omega = \frac{2 \pi v}{\lambda}$.Substituting this into the equation: $A \left( \frac{2 \pi v}{\lambda} \right) = v$.Solving for $A$: $A = \frac{v \lambda}{2 \pi v} = \frac{\lambda}{2 \pi}$.

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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