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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 |
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Answer» 1 cm 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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