Question

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

Answer 1

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