Question

A block of mass m is attached to a spring of spring constant k is free to oscillate with angular velocity $\omega$ in a horizontal plane without friction or clamping. It is pulled to a distance $x_0$ and pushed towards the center with a velocity $v_0$ at time t = 0. The amplitude of oscillations in terms of $\omega ,x_{0}and{v}_0$ is

• A. $\sqrt{\frac{v_0^2}{{\omega }^2}-x_0^2}$
• B. $\sqrt{{\omega }^2{v}_0^2+x_0^2}$
• C. $\sqrt{\frac{x_0^2}{{\omega }^2}+v_0^2}$
• D. $\sqrt{\frac{v_0}{{\omega }^2}+x_0^2}$

Answer 1

The correct option is D $\sqrt{\frac{v_0}{{\omega }^2}+x_0^2}$

Spring constant$=K$

Velocity $=\omega$

Distance $=x_0$

Velocity $=v_0$ at time $t=0$

Solution: The distance 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, $x=\frac{dx}{dt}$ (This is the formula of the SHM)

$V=-A\omega \sin (\omega t-\theta )$

At $t=0,x=x_0$

$x_0=A\cos \theta =\to \left(i\right)$

And $\frac{dx}{dt}=-v_0=A\omega \sin \theta$

$A\sin \theta =v_0/\omega \to \left(ii\right)$

Squaring and adding equations$\left(i\right)$ and $\left(ii\right)$ and we get,

$A^2({\cos }^2\theta +{\sin }^2\theta )=x_0^2+\left(\frac{v_0^2}{{\omega }^2}\right)$

$\therefore A=\sqrt{x_0^2+{\left(\frac{v_0}{\omega }\right)}^2}$

Hence, the amplitude of the resulting oscillation is

$A=\sqrt{x_0^2+{\left(\frac{v_0}{\omega }\right)}^2}$