Question

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

Answer 1

average acceleration of particle is $\frac{2\omega^2A}{\pi}$

the magnitude of average acceleration in half the time period for an simple harmonic motion is ....

standard equation of particle moving in SHM, x = Asinωt

after double differentiation we get,

d²x/dt² = a = -ω²Asinωt

for getting average acceleration during 0 to T/2

so, $a_{av}=\frac{\int\limits^{T/2}_0{a(t)}\,dt}{\int\limits^{T/2}_0dt}$

= $-\omega^2A\frac{\int\limits^{T/2}_0{sin\omega t}\,dt}{T/2}$

= $-\frac{2\omega^2A}{T}\left[\frac{-cos\omega t}{\omega}\right]^{T/2}_0$

as T = 2π/ω then, T/2= π/ω

= $\frac{4\omega^2 A}{2\pi}$

= $\frac{2\omega^2A}{\pi}$

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Answer 2

Step-by-step explanation:

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