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A string of length 'l' is fixed at both ends. It is vibrating in tis 3rd overtone with maximum amplitude 'a' the amplitude at a distance `l/3` from one end is

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The correct Answer is:
3
For third overtone `(4lambda)/(2) = l rArr 2lambda = l rArr lambda = (l)/(2)`
As `x = 0` is a node , `:. A_(s. omega) = A sin "(2pi)/('lambda) xx = a sin ((2pi)/((lambda)/(2))(1)/(3)) = a(sqrt(3))/(2)`
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