Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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 correctly shows the variation of acceleration, $a$ with time, $t$ ? ( $T =$ time period)

Graph showing positive cosine function

Graph showing negative sine function

Graph showing negative cosine function

Graph showing positive sine function

Step-by-Step Solution

Displacement Equation: The given equation for displacement is $X = A \cos(\omega t)$ .
Velocity: Velocity ( $v$ ) is the first derivative of displacement with respect to time: $v = \frac{dX}{dt} = \frac{d}{dt} (A \cos \omega t) = -A\omega \sin(\omega t)$
Acceleration: Acceleration ( $a$ ) is the first derivative of velocity (or second derivative of displacement) with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt} (-A\omega \sin \omega t) = -A\omega^2 \cos(\omega t)$
Analysis: The acceleration equation is $a = -A\omega^2 \cos(\omega t)$ . This is a negative cosine function. At $t=0$ , $\cos(0) = 1$ , so $a = -A\omega^2$ (maximum negative value). At $t=T/4$ , $\cos(\pi/2) = 0$ , so $a = 0$ . At $t=T/2$ , $\cos(\pi) = -1$ , so $a = +A\omega^2$ (maximum positive value). At $t=T$ , $\cos(2\pi) = 1$ , so $a = -A\omega^2$ .
Conclusion: The correct graph must start at a negative maximum, go through zero at $T/4$ , reach a positive maximum at $T/2$ , and return to a negative maximum at $T$ . This corresponds to a negative cosine curve (often labeled as Option C or 3 in standard exams like AIPMT).

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