Login/Sign Up
Home » A mass attached to a spring is free to oscillate, with angular velocity , in a horizontal plane without friction or damping. It is pulled to a distance and pushed towards the centre with a velocity at time Determine the amplitude of the resulting oscillations in terms of the parameters and [Hint: Start with the equation and note that the initial velocity is negative.]
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.]
CBSE | Class 11 | NCERT | Oscillations | Physics
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.]

The angular velocity of the spring be \omega

x=a \cos (\omega t+\theta)

At t=0, x=x_{0}

On Substituting these values in the above equation we get,

\mathrm{x}_{0}=\mathrm{A} \cos \theta-(1)

Velocity, v=d x / d t=-A \omega \sin (\omega t+\theta)

At t=0, v=-v_{0}

Again, on substituting these values in the above equation we get

-v_{0}=-A \omega \sin \theta

A \sin \theta=v_{0 / \omega}

Squaring and adding (1) and (2) we get,

A^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}

A=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}}

i

More Material