Question

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

• 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)

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