The oscillation of a body on a smooth horizontal surface is represented by the equation, X = A cos (ω t) where X = displacement at time t ω = frequency of oscillation Which one of the following graphs shows correctly the variation of a with t? Here a = acceleration at time t T = time period (a) (b) (c) (d)

Answer: (c). X = Acos ω t ⇒ a = (d²X)/(dt²) = -ω² Acos ω t. At t = 0: a = -ω² A (negative maximum); at t = T/2: a = +ω² A; at t = T: a = -ω² A again. This is an inverted cosine curve starting from the negative maximum. So the answer is (c).

The oscillation of a body on a smooth horizontal surface is… | NEET 2014 (AIPMT)

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Velocity & Acceleration in SHM

The oscillation of a body on a smooth horizontal surface is represented by the equation, X = A cos $(\omega t)$

where X = displacement at time t

$\omega$ = frequency of oscillation

Which one of the following graphs shows correctly the variation of a with t?

Here a = acceleration at time t

T = time period

Show answer and solution

Answer: (c)

$X = A\cos\omega t \Rightarrow a = \frac{d^2X}{dt^2} = -\omega^2 A\cos\omega t$ .
At $t = 0$ : $a = -\omega^2 A$ (negative maximum); at $t = T/2$ : $a = +\omega^2 A$ ; at $t = T$ : $a = -\omega^2 A$ again.
This is an inverted cosine curve starting from the negative maximum.
So the answer is (c).

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