A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is 50% of total energy? ( cos 4 5 ∘ = 2 1 )

Question

A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is 50% of total energy? ( cos 4 5 ∘ = 2 ​ 1 ​ )

TestHub · Accepted Answer

T = 6 sec 2 1 ​ m ω 2 ( A 2 − x 2 ) = 2 1 ​ ( 2 1 ​ m ω 2 A 2 ) A 2 − x 2 = 2 A 2 ​ ∴ x 2 = 2 A 2 ​ ⇒ x = 2 ​ A ​ 2 ​ A ​ = A sin ω t = A sin T 2 π ​ t 2 ​ 1 ​ = sin 6 2 π ​ t = sin 3 π ​ t sin 4 π ​ t = sin 3 π ​ t ∴ t = 4 3 ​ = 0.75 sec

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