Question

A body oscillates
with SHM according to the equation x = (5.0m)cos $\left[\left(2\pi rad{s}^{-1}\right)t+\pi /4\right]$ . At t = 1.5 s, its acceleration is

• A. -139.56 $m/s^2$
• B. 139.56$m/s^2$
• C. 69.78$m/s^2$
• D. -69.78$m/s^2$

Answer 1

The correct option is B 139.56$m/s^2$

We have,

$x=5\cos [2nt+\frac{\pi }{4}]$

$v=\frac{dx}{dt}=-5\sin [2nt+\frac{\pi }{4}]\left(2\pi \right)$

$r=\frac{dv}{dt}=-5(2\pi {)}^2\cos [2nt+\frac{\pi }{4}]$

At $t=1.5s$

$a=-5(2\pi {)}^2\cos (2\pi \times \frac{3}{2}+\frac{\pi }{4})$

$a=-5(2n{)}^2\cos \left(\frac{13\pi }{4}\right)$

$a=141.42$ if ${\pi }^2=10$

Option $B$ is correct