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Derive expressions for displacement, velocity and acceleration of a particle executing simple harmonic motion. |
Answer» Solution : Consdier the position of the particle P on a circle of radius A at some instant of TIME t as shown in Figure. Its displacement y at that instant of time t can be derive as follows In `DeltaOPN`, `"sin" theta=(ON)/(OP)` `rArr ON=OP sin theta ""...(1)` But `theta= omega t, ON=y and OP =A` `y=A sin omega t ""..(2)` The displacement y takes maximum value (which is EQUAL to A) when `sin omega t=1`. This maximum displacement from the mean position is known as AMPLITUDE (A) of the vibrating particle. for simple harmonic motion, the amplitude is constant. Velocity : The rate of change of displacement is velocity. TAKING derivative of equation (2) with RESPECT to time, we get `v=(dv)/(dt)=(d)/(dt) (A sin omega t)` For circular motion (of constant radius), amplitude A is a constant. Angular velocity `omega` is a constant for uniform circular motion. `:.v=(dy)/(dt) A omegac t ......(3)` Using trigonometry identity, `sin^(2)omegat+cos^(2)omega t=1` `rArr cos omega t= sqrt(1-sin^(2)omega t)` We get, `v=A omega sqrt(1-sin^(2) omegat)` Form equation (2), `"sin" omega t=(y)/(A)` `v= A omega sqrt(1-((y)/(A))^(2))"....(4)` From equation (4), when the displacement `y=0`, the velocity `v= omega A` (maximum) and for the maximum displacementy=A, the velocity v=0 (minimum). Acceleration : The rate of change of velocity is acceleration. `a=(dv)/(dt)=(d)/(dt) (A omega cos omegat)` `a=- omega^(2)A sin omegat=- omega^(2)y...(5)` `:. a=(d^(2)y)/(d^(2))=- omega^(2)y ""....(6)` |
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