$A$ mass attached to a spring is free to oscillate,with angular velocity $\omega$,in a horizontal plane without friction or damping. It is pulled to a distance $x_{0}$ and pushed towards the centre with a velocity $v_{0}$ at time $t=0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_{0}$ and $v_{0}$. [Hint: Start with the equation $x=A \cos (\omega t+\theta)$ and note that the initial velocity is negative.]

(A) The displacement equation for an oscillating mass is given by:
$x = A \cos (\omega t + \theta)$
Velocity is the derivative of displacement with respect to time:
$v = \frac{dx}{dt} = -A \omega \sin (\omega t + \theta)$
At $t = 0$,the displacement is $x = x_{0}$ and the velocity is $v = -v_{0}$ (since it is pushed towards the centre):
$x_{0} = A \cos \theta \quad \dots(i)$
$-v_{0} = -A \omega \sin \theta \implies A \sin \theta = \frac{v_{0}}{\omega} \quad \dots(ii)$
Squaring and adding equations $(i)$ and $(ii)$:
$A^2 \cos^2 \theta + A^2 \sin^2 \theta = x_{0}^2 + \left(\frac{v_{0}}{\omega}\right)^2$
$A^2 (\cos^2 \theta + \sin^2 \theta) = x_{0}^2 + \frac{v_{0}^2}{\omega^2}$
$A^2 = x_{0}^2 + \frac{v_{0}^2}{\omega^2}$
Therefore,the amplitude $A$ is:
$A = \sqrt{x_{0}^2 + \left(\frac{v_{0}}{\omega}\right)^2}$