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)

–3.54 m, 140 m s–2

Answer 2

The given equation of simple harmonic motion is

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

Compare the given equation with standard equation of SHM

$x(t) = A\cos(\omega t + \varphi)$

We get

$\omega = 2\pi s^{- 1}$

At $t = 1.5s$

Displacement $x(t) = 5\cos\left( 2\pi \times 1.5 + \frac{\pi}{4} \right)$

$= 5\cos\left( 3\pi + \frac{\pi}{4} \right) = - 5\cos\left( \frac{\pi}{4} \right)$

$= \lbrack\because\cos(3\pi + \theta) = - \cos\theta\rbrack$

$= - 5 \times 0.707m = - 3.54m$

Acceleration $a = - \omega^{2} \times$displacement

$= - (2\pi s^{- 1})^{2} \times ( - 3.54m) = 140ms^{- 2}$