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(A) The relationship between phase difference $\phi$ and path difference $x$ is given by $\phi = \frac{2 \pi}{\lambda} x$. First,calculate the wavelen

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(A) The relationship between phase difference $\phi$ and path difference $x$ is given by $\phi = \frac{2 \pi}{\lambda} x$. First,calculate the wavelength $\lambda$ using the formula $\lambda = \frac{v}{f}$,where $v = 330 \,m/s$ and $f = 50 \,Hz$. $\lambda = \frac{330}{50} = 6.6 \,m$. Now,rearrange the phase difference formula to solve for $x$: $x = \frac{\phi \lambda}{2 \pi}$. Substituting the values: $x = \frac{(\pi/3) 	imes 6.6}{2 \pi} = \frac{6.6}{6} = 1.1 \,m$. Thus,the distance between the two points is $1.1 \,m$.

$A$ sound wave is travelling with a frequency of $50 \,Hz$. The phase difference between two points in the path of the wave is $\frac{\pi}{3} \,rad$. The distance between those two points is (Velocity of sound in air $=330 \,m/s$) (in $m$)

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