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Vibrations of period 0.25 s propagate al...

Vibrations of period `0.25 s` propagate along a straight line at a velocity of `48 cm//s`. One second after the the emergence of vibrations at the initial point, displacement of the point, `47 cm` from it is found to be `3 cm`. [Assume that at initial point particle is in its mean position at `t=0` and moving upwards]. Then,

A

amplitude of vibrations is `6cm`

B

amplitude of vibrations is `3sqrt2cm`

C

amplitude of vibrations is `3cm`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`because` `omega =(2pi)/(T) =(2pi)/(0.25)`
`= 8pi rad//s`
`v=(omega)/(k)`
`:. k=(omega)/(v) =(8pi)/(0.48) = ((50)/(3))pi m^(-1)`
`y = A sin(omegat - kx)`
`=A sin (8pit-(50)/(3) pix)`
Put, `y = 3cm`, t= 1s,x=0.47m` showing we get `A = 6 cm`
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