A progressive wave of frequency 400 Hz is tr | vedclass

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Question

(A) Given: Frequency $(n) = 400 \ Hz$,Velocity $(v) = 336 \ m/s$,Phase difference $(\phi) = 60^{\circ} = \frac{\pi}{3} \ rad$.First,calculate the wave

Vedclass · Accepted Answer

$0.14$

Vedclass · Answer

(A) Given: Frequency $(n) = 400 \ Hz$,Velocity $(v) = 336 \ m/s$,Phase difference $(\phi) = 60^{\circ} = \frac{\pi}{3} \ rad$.First,calculate the wavelength $(\lambda)$ using the formula $v = n \lambda$:$\lambda = \frac{v}{n} = \frac{336}{400} = 0.84 \ m$.The relationship between path difference $(\Delta x)$ and phase difference $(\phi)$ is given by $\Delta x = \frac{\lambda}{2\pi} 	imes \phi$.Substituting the values:$\Delta x = \frac{0.84}{2\pi} 	imes \frac{\pi}{3} = \frac{0.84}{6} = 0.14 \ m$.Therefore,the two points are $0.14 \ m$ apart.

$A$ progressive wave of frequency $400 \ Hz$ is travelling with a velocity $336 \ m/s$. How far apart are the two points which are $60^{\circ}$ out of phase (in $m$)?

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