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A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode.Amplitude at the Centre of the string is 4 mm. Distance between the two points having amplitude 2 mm is

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$50 c m$

$72 c m$

$1 m$

$60 c m$

answer is A.

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Detailed Solution

$\frac{λ}{2} = 1.5 m ∴ λ = 3.0 m = k = \frac{2 π}{λ} = (\frac{2 π}{3}) m^{- 1}$

Let us take antinode at x=0, then

$A_{x} = A_{\max} \cos k x ∴ (2 m m) = (4 m m) \cos (\frac{2 π}{3}) x ∴ (\frac{2 π}{3}) x = \frac{π}{3} o r x = 0.5 m$

The asked distance is 2x or 1.0 m

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