Velocity of sound waves in air is 330 m/s. F | vedclass

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(C) Given: Velocity of sound $v = 330 \ m/s$,path difference $x = 40 \ cm = 0.4 \ m$,and phase difference $\phi = 1.6 \pi$.We know the relationship be

Vedclass · Accepted Answer

$660$

Vedclass · Answer

(C) Given: Velocity of sound $v = 330 \ m/s$,path difference $x = 40 \ cm = 0.4 \ m$,and phase difference $\phi = 1.6 \pi$.We know the relationship between phase difference and path difference is given by $\phi = \frac{2 \pi}{\lambda} \cdot x$.Substituting the values: $1.6 \pi = \frac{2 \pi}{\lambda} 	imes 0.4$.$1.6 = \frac{0.8}{\lambda} \implies \lambda = \frac{0.8}{1.6} = 0.5 \ m$.Using the wave equation $v = f \lambda$,we find the frequency $f = \frac{v}{\lambda}$.$f = \frac{330}{0.5} = 660 \ Hz$.

Velocity of sound waves in air is $330 \ m/s$. For a particular sound wave in air,a path difference of $40 \ cm$ is equivalent to a phase difference of $1.6 \pi$. The frequency of this wave is: (in $Hz$)

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