A mass attached to a spring is free to oscillate, with angular ve… - Vidyadip

Question

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 $\upsilon_0$ at time $t = 0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_0$ and $\upsilon_0.$ [Hint: Start with the equation $\text{x}=\text{a}\cos(\omega\text{t}+\theta)$ and note that the initial velocity is negative.]

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Answer

The displacement rquation for an oscillating mass is given by:$\text{x}=\text{A}\cos(\omega\text{t}+\theta)$
where,
A is the amplitude
x is the displacement
$\theta$ is the phase constant
Velcoity, $\text{v}=\frac{\text{dx}}{\text{dt}}=-\text{A}\omega\sin(\omega\text{t}+\theta)$
At, $t = 0, x = x_0$
$\text{x}_0=\text{A}\cos\theta=\text{x}_0\ .....(\text{i})$
and, $\frac{\text{dx}}{\text{dt}}=-v_0=\text{A}\omega\sin\theta$
$\text{A}\sin\theta=\frac{v_0}{\omega}\ ....(\text{ii})$
Squaring and adding equations (i) and (ii), we get:
$\text{A}^2(\cos^2\theta+\sin^2\theta)=\text{x}_0^2+\Big(\frac{v_0^2}{\omega^2}\Big)$
$\therefore\ \text{A}=\sqrt{\text{x}_0^2+\Big(\frac{v_0}{\omega}\Big)^2}$
Hence, the amplitude of the resulting oscillation is $\text{x}_0^2+\Big(\frac{v_0}{\omega}\Big)^2.$

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