The velocity of sound waves in air is 330 ; m | vedclass

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Question

(C) The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by: $\Delta \phi = \frac{2\pi}{\lambda} 	imes

Vedclass · Accepted Answer

$660$

Vedclass · Answer

(C) The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by: $\Delta \phi = \frac{2\pi}{\lambda} 	imes \Delta x$. Given $\Delta \phi = 1.6 \pi$ and $\Delta x = 40 \; cm = 0.4 \; m$. Substituting these values: $1.6 \pi = \frac{2\pi}{\lambda} 	imes 0.4$. Solving for wavelength $(\lambda)$: $\lambda = \frac{2 	imes 0.4}{1.6} = \frac{0.8}{1.6} = 0.5 \; m$. Using the wave equation $v = f \lambda$,where $v = 330 \; m/s$ and $\lambda = 0.5 \; m$: $330 = f 	imes 0.5$. Therefore,$f = \frac{330}{0.5} = 660 \; Hz$.

The 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 ... $Hz$.

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