Question

The position \(x(t)\) of a particle, at constant \(\omega\), is described by \(\dfrac{d^{2}x}{dt^{2}}=-\omega^{2}x\) with initial conditions \(x(0)=1\) and \(\left.\dfrac{dx}{dt}\right|_{t=0}=0\). The position of the particle at \(t=\dfrac{3\pi}{\omega}\) is (in integer).

Hint:

For \(x''+\omega^2 x=0\), use \(x=A\cos\omega t + B\sin\omega t\) and apply initial conditions.
Remember \(\cos(\pi)= -1\), \(\cos(2\pi)=1\), \(\cos(3\pi)=-1\).

Answer 1

Correct Answer: -1

Step 1: The ODE is simple harmonic motion; the solution satisfying \(x(0)=1,\; \dot x(0)=0\) is \[ x(t)=\cos(\omega t). \] Step 2: Evaluate at \(t=\dfrac{3\pi}{\omega}\): \[ x\!\left(\frac{3\pi}{\omega}\right)=\cos(3\pi)=-1. \]