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Show that for a particle executing simple harmonic motion average value of kinetic energy is equal to the average value of potential energy. |
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Answer» Solution :Suppose a particle of mass m exectues SHM of time PERIOD T. The DISPLACEMENT of the particle at any instant t is given by `y=A sin OMEGA t""....(1)` Velocity `v=(dy)/(dt)=(d)/(dt)(A sin omega t)` `=A omega cos omega t= omega A cos omega t` `:. {:("Kinetic"),("energy"):}}E_(k)=(1)/(2)mv^(2)` `=(1)/(2) m omega^(2)A^(2) cos^(2)omega t""....(2)` `:. {:("Potential"),("energy"):}}E_(P)=(1)/(2)m^(2) v^(2)` `=(1)/(2) m omega^(2)A^(2) sin^(2) omega t""....(3)` `:.` Average K.E for one period of oscillation, `E_(k_(av))=(1)/(T_(0)) E_(k)dt` `=(1)/(T) overset(T) UNDERSET(0)int (1)/(2) m omega^(2)A^(2)cos ^(2) omega t dt` `=(1)/(2T)m omega^(2)A^(2) overset(T) underset(0)int ((1-cos2 omega t))/(2)dt` `K_(k_(av))=(1)/(4T)m omega^(2)A^(2)[t+(sinj 2 omega t)/(2 Omega)]_(0)^(T)` `K_(k_(av))=(1)/(4T)m omega^(2)A^(2)[T+0]` `k_(k_(av))=(1)/(4T)m omega^(2)A^(2)(T)` `E_(K_(av))=(1)/(4)m omega^(2)A^(2) ""...(4)` Average potential energy over a period of oscillation is `E_(p_(av))=(1)/(T)overset(T) underset(0)int E_(rho)dt` `=(1)/(T)overset(T) underset(0)int m omega^(2)A^(2)sin^(2) omega t dt` `=(1)/(2)m omega^(2)A^(2)A^(2)overset(T) underset(0)int ((1- cos 2 omega t))/(2)dt` `E_(p_(av))=(1)/(4) m omega^(2)A^(2)[-t-(sin 2 omegat)/(2 omega)]_(0)^(T)` `=(1)/(4T) m omega^(2)A^(2)[T-0]` `=(1)/(4T)m omega^(2)A^(2)(T)` `E_(p_(av))=(1)/(4T) m omega^(2)A^(2) ""....(5)` Clearly from equation (4) and (5) `E_(K_(av))=E_(p_(av))` `:. {:("Average"),("value of K.E"):}}= {{:("Average"),("value of P.E"):}` |
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