Question

The $ SHM $ of a particle is given by $ x\left(t\right)=5\,cos\left(2\pi t+\frac{\pi}{4}\right) $ (in $ MKS $ units).
Calculate the displacement and the magnitude of acceleration of the particle at $ t =1.5\,s $ .

• A. $ -3.0\, m $ , $ 100 \,m/s^2 $
• B. $ +2.54\,m $ , $ 200 \,m/s^2 $
• C. $ -3.54\,m $ , $ 140 \,m/s^2 $
• D. $ +3.55\, m $ , $ 120\, m/s^2 $

Answer 1

Answer: (C)

Displacement $x \left(t\right)=5\backslash,cos \left(2\pi\times\frac{3}{2}+\frac{\pi}{4}\right)$
$5\,cos \left[\frac{13 \pi}{4}\right]=-3.5\,m$
$y=5\,cos \left(2\pi t+\frac{\pi}{4}\right)$
$\therefore V=-10\pi\,sin \left(2\pi t+\frac{\pi}{4}\right)$
$\therefore $ acceleration $= -20\pi^{2} cos \left(2\pi t+\frac{\pi}{4}\right)$
$=-20\pi^{2}\, cos \left(2\pi\times\frac{3}{2}+\frac{\pi}{4}\right)$
$=20\pi^{2}\,cos \frac{13\pi}{4}$
$=140 m/ s^{2}$