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Q.

A string of length 1.5 m with its two ends clamped is vibrating in the fundamental mode. The amplitude at the centre of the string is 4 mm. Find the minimum distance between the two points that have amplitude 2 mm.

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a

1 m

b

75 cm

c

60 cm

d

50 cm

answer is A.

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Detailed Solution

The string is vibrating in its fundamental mode. Hence the wavelength= 1.5 x 2= 3m

$y = A \sin k x \cos ω t 2 = 4 \sin (k x) \cos (ω t) \frac{1}{2} = \sin k x \cos ω t A t t = 0, k x = \frac{π}{6}, \frac{5 π}{6} \frac{2 π}{λ} x = \frac{π}{6}, \frac{5 π}{6} p h a s e d i f f e r e n c e = \frac{2}{3} π Thus separation = \frac{λ}{3} = 1 m$

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