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An oscillator executes SHM is given by y = A Sinωt + BCosωt, Find the acceleration of SHM when ωt= \(\frac{\pi}{2}\)
1. -ω2Sinωt
2. ω3Cos3ω t
3. -Aω2
4. +ω2Sinωt

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Correct Answer - Option 3 : -Aω2


  • Simple harmonic motion is the repetitive motion of a body back and forth about an equilibrium point. \
  • The force which executes this motion is given by

⇒ F = -kx

Where k = spring constant, x = Position

  • The position of a simple harmonic motion is given by

⇒ x = A Cos (ωt + Φ)

Differentiating the above equation the velocity can be written as

⇒ V = -AωSin(ωt + ϕ)

Differentiating the above equation we get the acceleration as

⇒ a = -ω2Cos(ωt + ϕ)

  • in a simple harmonic motion, the acceleration is directed towards the center


Given - y = A Sinωt+BCosωt

  • The velocity is given by

\(\Rightarrow V = \dfrac{dy}{dt} = \dfrac{d(A Sinω t+B Cosω t)}{dt}\)

⇒ V = A ω Cos ωt -Bω Sin ωt

Again differentiating the above equation with respect to t

\(⇒ a = \dfrac{dV}{dt} = \dfrac{d(A ω Cos ω t - B ω Sinω t)}{dt} = - Aω \times ω Sinω t - Bω\times ω Cosω t\)

⇒ a = - Aω2Sin ωt -B ω2Cosωt 

Substituting the value \(\omega t = \dfrac{\pi}{2}\)

⇒ a = -A ω2

  • Hence, option 3 is the answer

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