Question

A particle executing SHM according to the equation $x=5cos[2\pi t+\frac{\pi }{4}]$ in SI units. The displacement and acceleration of the particle at t = $1.5$s is:

• 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

The correct option is C $-3.54m,140m{s}^{-2}$

The given equation of simple harmonic motion is
$x\left(t\right)=5cos[2\pi t+\frac{\pi }{4}]$
Compare the given equation with standard equation of SHM
$x\left(t\right)=Acos(\omega t+\varphi )$
we get, $\omega =2\pi s^{-1}$
At $t=1.5s$
Displacement, $x\left(t\right)=5cos[2\pi \times 1.5+\frac{\pi }{4}]=5cos[3\pi +\frac{\pi }{4}]=-5cos\left[\frac{\pi }{4}\right]=-5\times 0.707m=-3.54m$
Acceleration, $a=-{\omega }^2\times displacement=-(2\pi s^{-1}{)}^2\times (-3.54m)=140m{s}^{-2}$