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Q. A wave equation is $ y=0.1\sin [100\pi t-kx] $ and wave velocity is 100 m/s, its wave number is equal to

ManipalManipal 2008Waves

Solution:

The standard equation of wave is $ y=a\sin (\omega t-kx) $ .. (i)
where $ a $ is amplitude, $ \omega $ the angular velocity and $ x $ the displacement at instant $ t $ .
Given equation is $ y=0.1\,\,\sin (100\pi t-kx) $ .. (ii)
Comparing Eq. (i) with Eq. (ii), we get
$ \omega =100\pi $
$ \therefore $ Wave number
$=\frac{\omega }{v}=\frac{100\pi }{100}=\pi \,\,{{m}^{-1}} $