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A particle on a stretched string supporting a travelling wave, takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed.

Text Solution

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The correct Answer is:
50hz, 4.0 cm, 2.0 m/s
Given that time period
`T = 5 xx 4 = 20 m//s`
implies frequency `n = (l)/(t) = (1000)/(20) = 50 Hz`
`(lambda)/(2)` = 2 cm
`lambda` = 4cm
Wave speed `v = nlambda = 50 xx 0.04 = 2m//s`
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