Question

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

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

Answer 1

Answer: (C)

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