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

Question

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)

Accepted Answer

1.  **Displacement Equation:** The given equation for displacement is $X = A \cos(\omega t)$.
2.  **Velocity:** Velocity ($v$) is the first derivative of displacement with respect to time:
    $$ v = \frac{dX}{dt} = \frac{d}{dt} (A \cos \omega t) = -A\omega \sin(\omega t) $$
3.  **Acceleration:** Acceleration ($a$) is the first derivative of velocity (or second derivative of displacement) with respect to time:
    $$ a = \frac{dv}{dt} = \frac{d}{dt} (-A\omega \sin \omega t) = -A\omega^2 \cos(\omega t) $$
4.  **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 $a = -A\omega^2$.
5.  **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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