The distance between two consecutive points w | vedclass

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(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula: $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.Give

Vedclass · Accepted Answer

$9.6$

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(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula: $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.Given $\Delta \phi = 45^{\circ} = \frac{\pi}{4} 	ext{ radians}$ and $\Delta x = 4.0 	ext{ m}$.Substituting these values: $\frac{\pi}{4} = \frac{2\pi}{\lambda} 	imes 4.0$.Solving for wavelength $\lambda$: $\lambda = 8 	imes 4.0 = 32 	ext{ m}$.The wave velocity $v$ is given by $v = f \lambda$,where $f = 300 	ext{ Hz}$.$v = 300 	ext{ Hz} 	imes 32 	ext{ m} = 9600 	ext{ m/s}$.Converting to $	ext{km/s}$: $v = \frac{9600}{1000} 	ext{ km/s} = 9.6 	ext{ km/s}$.

The distance between two consecutive points with phase difference of $45^{\circ}$ in a wave of frequency $300 \text{ Hz}$ is $4.0 \text{ m}$. The velocity of the travelling wave is (in $\text{km/s}$):

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