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  • A string of 1m with its two ends clamped is vibrating in fundamental mode. Amplitude the centre of string is 8mm. Miniumum distance between the two points having amplitude 4mm is<div class='qna

A string of 1m with its two ends clamped is vibrating in fundamental mode. Amplitude the centre of string is 8mm. Miniumum distance between the two points having amplitude 4mm is

Option: 1

1m


Option: 2

67 cm


Option: 3

75cm


Option: 4

33 cm


Answers (1)

best_answer

\lambda = 2l = 2m

Equation of standing wave

y = 2A \sin kx \cos \omega t

Amplitude = 2A\sin k x

4mm = 8 \sin\left ( \frac{2\pi}{\lambda}x \right )

\frac{1}{2}= \sin\left ( \frac{2\pi}{\lambda}x \right )

\frac{2\pi x}{ \lambda}=30^{\circ}=\frac{\pi}{6}\Rightarrow x_1=\frac{2}{12}=\frac{1}{6}=0.166m

or

\frac{2\pi x}{ \lambda}=150^{\circ}=\frac{5\pi}{6}\Rightarrow x_2=\frac{10}{12}=\frac{1}{6}=0.833m

\Delta x=\frac{5}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}=0.67m

Posted by

SANGALDEEP SINGH

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