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A mass attached to a spring is free to oscillate, with angular velocity $\omega,$ in a hortzontal plane without friction or damping. It is pulled to a distance $x_{0}$ and pushed towards the centre with a velocity $v_{ o }$ at time $t=0 .$ Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_{0}$ and $v_{ o } .$ [Hint: Start with the equation $x=a \cos (\omega t+\theta)$ and note that the initial velocity is negative.]
Solution
The displacement equation for an oscillating mass is given by:
$x=A \cos (\omega t+\theta)$
Where,
$A$ is the amplitude $x$
is the displacement $\theta$
is the phase constant
Velocity, $v=\frac{d x}{d t}=-A \omega \sin (\omega t+\theta)$
At $t=0, x=x_{0}$
$A \cos \theta=x_{0} \ldots(i)$
And, $\frac{d x}{d t}=-v_{0}=A \omega \sin \theta$
$A \sin \theta=\frac{v_{0}}{\omega} \ldots(i i)$
Squaring and adding equations ( $i$ ) and ($ ii $), we get
$A^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=x_{0}^{2}+\left(\frac{v_{0}^{2}}{\omega^{2}}\right)$
$\therefore A=\sqrt{x_{0}^{2}+\left(\frac{v_{0}}{\omega}\right)^{2}}$
Hence, the amplitude of the resulting oscillation is $\sqrt{x_{0}^{2}+\left(\frac{v_{0}}{\omega}\right)^{2}}$
