Correct Answer – `y = a sin omegat` , `v = delomegat`
Kinetic energy, `E_(k) = (1)/(2)mv^(2) = E_(k) = (1)/(2)ma^(2)omega^(2)cos^(2)omegat`
Potential energy, kinectic energy, `E_(k) = (1)/(2)mv^(2) = E_(k) = (1)/(2)ma^(2)omega^(2)cos^(2)omegat`
Average kinetic energy over one cycle
`= (1)/(T)(int^(T)_(0)) E_(k)dt`
`= (1)/(T)(int_(0)^(T))(1)/(2)ma^(2)omega^(2)cos^(2)omegatdt`
`= (1)/(2T)ma^(2)omega^(2)int_(0)^(T)cos^(2)omegatdt`
`= (1)/(2T)ma^(2)omega^(2)int_(0)^(T)(1)/(2)(1 + cos2omegat)dt | or {:(cos2theta = 2cos^(2)theta – 1),(2cos^(2)theta = 1 + cos2theta):}` = (1)/(4Tma^(2)omega^(2)[int_(0)^(T)1dt+int_(0)^(T)cos2omegat dt]` a
`= (ma^(2)omega^(2))/(4T)[T+|(sin2omegat)/(2omega)|_(0)^(T)]`
`= (ma^(2)omega^(2))/(4T)[T + (1)/(2omega)(0 – 0)} = (1)/(4)momega^(2)a^(2)`
Average potential energy over cycle
`(1)/(T)int_(0)^(T)E_(p)dt = (1)/(T)int_(0)^(T)(1)/(2)momega^(2)a^(2)sin^(2)omegat`
`= (1)/(2T)momega^(2)a^(2)int_(0)^(T)sin^(2)omegatdt`
`(1)/(2T)momega^(2)a^(2)int_(0)^(T)(1)/(2)(1 – cos2theta)dt`
`|or{:(cos2theta=1-2sin^(2)theta),(2sin^(2)theta=1cos2theta):}=(1)/(4T)momega^(2)a^(2)[int_(0)^(T)cos2omegatdt]`
`= (1)/(4) momega^(2)a^(2)`