Question

A particle performing simple harmonic motion about O with amplitude A and time period T. The magnitude of its acceleration at t=T/8s after the particle reaches that extreme position would be -

• A. $\frac{{4\pi }^{2}A}{\sqrt{2}{T}^2}$
• B. $\frac{{4\pi }^{2}A}{T^2}$
• C. $\frac{{2\pi }^{2}A}{\sqrt{2}{2T}^2}$
• D. None of these

Answer 1

Answer: (A)

$\frac{{4\pi }^{2}A}{\sqrt{2}{T}^2}$

Let at $t=0$ the particle is at extreme position, then the equation of SHM can be written as
$x=A\cos \left(\omega t\right)=A\cos \left(\frac{2\pi }{T}t\right)$
At $t=T/8$,
$x=A\cos \frac{\pi }{4}=\frac{A}{\sqrt{2}}$
Acceleration $=-{\omega }^{2}x=-{\left(\frac{2\pi }{T}\right)}^2\times \frac{A}{\sqrt{2}}$
Magnitude of acceleration $=\frac{4{\pi }^{2}A}{\sqrt{2T^2}}$