A simple harmonic progressive wave of friction $5 h z$ is traveling along the positive direction with a velocity of $40 m / s$ . Calculate the phase difference between two points separated by a distance of $0.8 m$ .

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Solution

$f = 5 H z$
$v = 40 m / s$
as, $λ = \frac{v}{f}$ [wavelength]
$λ = \frac{40}{5} = 8 m$
as, $λ$ path difference means $2 π$ phase difference.
$∴ 0.8$ path difference= $\frac{2 π}{10}$ phase difference
So, phase difference, $\frac{π}{5}$ rad

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A simple harmonic progressive wave of friction 5hz is traveling along the positive direction with a velocity of 40m/s. Calculate the phase difference between two points separated by a distance of 0.8m.

f=5Hzv=40m/sas, λ=vf [wavelength]λ=405=8mas, λ path difference means 2π phase difference.∴0.8 path difference=2π10 phase differenceSo, phase difference, π5rad

A simple harmonic progressive wave of friction $5 h z$ is traveling along the positive direction with a velocity of $40 m / s$ . Calculate the phase difference between two points separated by a distance of $0.8 m$ .

$f = 5 H z$
$v = 40 m / s$
as, $λ = \frac{v}{f}$ [wavelength]
$λ = \frac{40}{5} = 8 m$
as, $λ$ path difference means $2 π$ phase difference.
$∴ 0.8$ path difference= $\frac{2 π}{10}$ phase difference
So, phase difference, $\frac{π}{5}$ rad