A particle executes simple harmonic motion according to x = a sin(50t+π / 3), where x is in centimetres and t is in seconds. Starting from t = 0, let t₁ be the first time at which the particle comes to rest and t₂ be the first time at which its acceleration becomes zero. What are t₁ and t₂, respectively?
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A particle executes simple harmonic motion according to x = a sin(50t+π / 3), where x is in centimetres and t is in seconds. Starting from t = 0, let t₁ be the first time at which the particle comes to rest and t₂ be the first time at which its acceleration becomes zero. What are t₁ and t₂, respectively?

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G1 Physics · Accepted Answer

The phase of the SHM is φ = 50t+π ÷ 3. The particle comes to rest when its velocity is zero. For x = a sinφ, velocity is proportional to cosφ. Therefore the first time after t = 0 when v = 0 occurs at φ = π ÷ 2. Hence, 50t₁+π ÷ 3 = π ÷ 2. Thus, 50t₁ = π ÷ 2-π ÷ 3 = π ÷ 6. Therefore, t₁ = π ÷ 300 s. Acceleration in SHM is a(SHM) = -ω²x. Thus the acceleration is zero when x = 0. Starting from phase π ÷ 3, the next zero of x occurs at φ = π. Therefore, 50t₂+π ÷ 3 = π. So, 50t₂ = 2π ÷ 3. Hence, t₂…