Question

The magnitude of average acceleration in half time period in a simple harmonic motion is:

• A. $\frac{2{\omega }^{2}A}{T}$
• B. $A^2/2T$
• C. $2A/2\omega$
• D. Zero

Answer 1

Answer: (A)

$\frac{2{\omega }^{2}A}{T}$

Let displacement function is $x=A\sin \omega t$

$A$= amplitude

$\omega$= angular frequency of oscillation.$=\frac{2\pi }{T}$

$T$= time period.

The average acceleration over half time period

$a_{av}=\frac{d^{2}x}{{dt}^2}=-{\omega }^{2}A\sin \omega t$

$a_{av}=\frac{{\int }_0^{\frac{T}{2}}a\left(t\right)dt}{{\int }_0^{\frac{T}{2}}dt}$

$=\frac{2}{T}(-{\omega }^{2}A){\int }_0^{\frac{T}{2}}\sin \left(\omega t\right)dt$

$=\frac{2}{T}(-{\omega }^{2}A)\frac{1}{\omega }[-\cos \left(\omega t\right){]}_0^{\frac{T}{2}}$

$=\frac{2\omega A}{T}[-(-\cos \left(\frac{\omega t}{2}\right))+\cos 0]$

$=-\frac{2\omega A}{T}[\cos \left(\frac{2\pi }{2T}\right)+1]$

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

So, the magnitude is $\frac{2{\omega }^{2}A}{T}$