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 intial point, displacement of the point, `47 cm` from it is found to be `3 cm`. [Assume that at intial point particle is in its mean position at `t=0` and moving upwards]. Then,

Options

A. amplitude of vibrations is `6cm`
B. amplitude of vibrations is `3sqrt2cm`
C. amplitude of vibrations is `3cm`
D. None of these

Answer

A

Explanation

`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`