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

\frac{2 A ω^{2}}{π}

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\frac{A ω^{2}}{2 π}

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\frac{A ω^{2}}{\sqrt{2 π}}

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Solution

The correct option is C $\frac{A ω^{2}}{2 π}$
Average acceleration is defined as $\frac{Δ v e l o c i t y}{Δ t i m e} a_{v} = \frac{V_{m a x}}{T} = \frac{V_{m a x}}{2 π} \times ω = \frac{A ω \times ω}{2 π} = \frac{A ω^{2}}{2 π}$

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

The correct option is C Aω22πAverage acceleration is defined as ΔvelocityΔtimeav=VmaxT=Vmax2π×ω=Aω×ω2π=Aω22π

The correct option is C $\frac{A ω^{2}}{2 π}$
Average acceleration is defined as $\frac{Δ v e l o c i t y}{Δ t i m e} a_{v} = \frac{V_{m a x}}{T} = \frac{V_{m a x}}{2 π} \times ω = \frac{A ω \times ω}{2 π} = \frac{A ω^{2}}{2 π}$