An oscillator executes SHM is given by y = A Sinωt + BCosωt, Find the acceleration of SHM when ωt= $\frac{\pi}{2}$

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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

CONCEPT

• 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

CALCULATION:

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