Question

The magnitude of average acceleration is half time period for equilibrium position is a simple harmonic motion is

Answer 1

Answer:

$a=-\frac{2Aw^{2}}{\pi }$

Step-by-step explanation:

average acceleration is given by ∫adt÷∫dt

let equation of SHM be x=A sinwt

therefore equation of  acceleration after differentiating twice a=-Aw2sinwt

average acceleration =$\int_{0}^{T/2}-Aw^{2}sinwt$

$\int_{0}^{T/2}dt$

$=\frac{-2Aw^{2}}{T}\frac{1}{w}\int_{0}^{T/2}-coswt$

$=\frac{2Aw}{T}[coswT/2-cos0]$

$a=\frac{2\pi }{w}$

$a=-\frac{2Aw^{2}}{\pi }$