1.

A particle in simple harmonic motion is described by the displacement functionx(t) = A cos(omegat +theta).If the initial (t = 0) position of the particle is 1 cm, its initial velocity is it cm/s, and its angular speed is pi radian per second then its amplitude is

Answer»

1 cm
`SQRT(2)` cm
2 cm
2.5 cm

Solution :Given, `x(t) = A cos (omega t+ phi)`
At t=0, I = A cos `(0 + theta) = A cos theta`
`rArr cos theta =1/A`...........(i)
Velocity of PARTICLE `=(DX)/(dt) =-AOMEGA. Sin(omegat + phi)`
`pi =-A.pi sin theta rArr sin theta =1/A`............... (ii)
By SQUARING and adding equation (i) and (ii) we get
`cos^(2) theta + sin^(2) theta =1/A^(2) + 1/A^(2) =2/A^(2)`
`therefore A^(2) =2` or `A = sqrt(2) cm`


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