Question

A particle is moving along the x-axis such that its acceleration is proportional to the displacement from the equilibrium position and they are in the same direction. The displacement $x(t)$ is given by

• A. $\sin \omega t, \omega > 0$
• B. $\sin \omega t + \cos \omega t, \omega > 0$
• C. $e^{\omega t}, \omega > 0$
• D. $e^{\omega t} + \sin \omega t, \omega > 0$
• E. $e^{\omega_1 t} + e^{-\omega_2 t}, \omega_1 \text{ and } \omega_2 > 0$

Hint:

If acceleration and displacement have the same sign ($a = \omega^2 x$), the motion is non-oscillatory and involves exponential growth or decay. If they have opposite signs ($a = -\omega^2 x$), the motion is oscillatory (SHM).

Answer 1

The Correct Option is C

Concept:
Acceleration ($a$) is the second derivative of displacement ($x$) with respect to time ($t$). The problem states that $a \propto x$ and they are in the same direction. [itemsep=8pt]
• Mathematical Form: $\frac{d^2x}{dt^2} = kx$, where $k$ is a positive constant ($k = \omega^2$).
• Comparison: This is the opposite of Simple Harmonic Motion, where $a = -\omega^2 x$. In this case, the displacement grows exponentially rather than oscillating.

Step 1: Solve the differential equation.
The equation is $\frac{d^2x}{dt^2} - \omega^2 x = 0$. The general solution for this linear differential equation is: \[ x(t) = C_1 e^{\omega t} + C_2 e^{-\omega t} \]

Step 2: Evaluate the options.
[itemsep=6pt]
• Options (A) and (B) represent SHM ($a = -\omega^2 x$).
• Option (C) $e^{\omega t}$ is a specific solution of the differential equation where $C_2 = 0$.
• Differentiating $x = e^{\omega t}$ twice gives $a = \omega^2 e^{\omega t} = \omega^2 x$, which satisfies the condition.