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A particle executes SHM of period 8 seconds. After what time of its passing through the mean position will the energy be half kinetic and half potential ? |
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Answer» Here, `T=8s,` As, `PE=KE` `:. (1)/(2)ky^(2)=(1)/(2)k(A^(2)-y^(2)) or y^(2)=A^(2)-y^(2)` `y=(A)/(sqrt(2))` Now, `y=Asin omega t =A sin ((2pi)/(T))t` `:. (A)/(sqrt(2))=A sin ((2pi)/(8))t` or `sin ((pit)/(4))=(1)/(sqrt(2))=sin ((pi)/(4))` or ` (pit)/(4)=(pi)/(4) or t=1s` |
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