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प्रश्न
A string 1m long is fixed at one end. The other end is moved up and down with frequency of 15 Hz. Due to this, a stationary wave with four complete loops gets produced on the string. Find the speed of the progressive wave which produces the stationary wave.
[Hint: Remember that the moving end is an antinode.]
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उत्तर
Data: L = 1 m, n = 15 Hz.
Only one end of the string is fixed. As a result, at the free end, an antinode will form. Hence, with four and a half loops on the string, the string length is
L = `λ/4+4(λ/2)=9/4λ`
∴ λ = `("4L")/9=4/9xx1=4/9`m
v = n λ
∴ Speed of the progressive wave
v = `15xx4/9=60/9` = 6.667 m/s
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