Question

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

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

Answer 1

The correct option is

C

$139.56m/s^2$

$x=5\cos (2\pi t+\frac{\pi }{4})$

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

$a=\frac{d^{2}x}{d{t}^2}=-5(2\pi {)}^2\cos (2\pi t+\frac{\pi }{4})$

Put $t=1.5$

$a=+139.56m/s^2$

$\cos (3\pi +\frac{\pi }{4})=\frac{-1}{\sqrt{2}}$

So option (B).