First slide

Simple hormonic motion

Question

The displacement equation of a particle is is $x = (3 sin 2 t + 4 cos 2 t) c m$ . Then magnitude of maximum acceleration of the particle is

Moderate

A

60 $\frac{c m}{s^{2}}$
B

10 $\frac{c m}{s^{2}}$
C

20 $\frac{c m}{s^{2}}$
D

40 $\frac{c m}{s^{2}}$

Solution

$x = 3 sin 2 t + 4 cos 2 t = \sqrt{3^{2} + 4^{2}} sin (2 t + {tan}^{- 1} 4 / 3)$
$∴ A = Amplitude = 5 c m$
$ω = 2 r a d / sec$
$∴$ Maximum acceleration $= ω^{2} A = {(2)}^{2} \times 5 c m / s^{2} = 20 c m / s^{2}$

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