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Length of a stretched wire is 2m. It is ...

Length of a stretched wire is 2m. It is oscillating in its fourth overtone mode. Maximum amplitude of oscillations is 2mm. Find amplitude of oscillation at a distance of 0.2m from one fixed end.

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The correct Answer is:
B
Fourth overtone mode means five loops.

`:. l= 5(lambda/2)`
or `lambda = (2l)/5`
` = (2xx2)/5 = 0.8m `
`k=(2pi)/lambda = (2pi)/0.8
`=(2.5 pi) m^(-1)`
` Now, P is a node. So, take x = 0 at P. `
Then, at a distance x from P,
`A_x = A_(max) sin kx`
` = (2mm) sin (2.5 pi) (0.2)`
`= (2mm) sin (pi/2)`
`=2mm` .
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