A particle of mass ' m ' is executing simple harmonic motion about its mean position. If ′ A ′ is the amplitude and ' T ' is the period of S.H.M., then the total energy of the part

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A particle of mass ' m ' is executing simple harmonic motion about its mean position. If ′ A ′ is the amplitude and ' T ' is the period of S.H.M., then the total energy of the particle is

TestHub · Accepted Answer

x = A sin ω t , , = d t d x ​ = ω A cos ω t T ⋅ E = P ⋅ E + K ⋅ E , P 0 ​ E = 2 1 ​ k x 2 , K ⋅ E = 2 1 ​ m v 2 T. E = 2 1 ​ × k × A 2 sin 2 ω t + 2 1 ​ m ω 2 A 2 cos 2 ω t , ω = m R ​ ​ , km v 2 T. E = 2 1 ​ m ω 2 A 2 sin 2 ω t + 2 1 m ω 2 A 2 c c 2 ω t ​ ⇒ 2 m ω 2 A 2 ​ = T 2 2 m π 2 A 2 ​

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