Question

A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x₀ and v₀. [Hint: Start with the equation x = acos (ωt+θ) and note that the initial velocity is negative.]

Answer 1

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

θ is the phase constant

Velocity, `v = (dx)/(dt) = -Aomega sin(omega t + theta)`

At t = 0, x = x₀

x₀ = Acosθ = x₀ … (i)

And, `(dx)/(dt) = -v_0 = Aomegasin theta`

`A sin theta = v_0/omega` ...(ii)

Squaring and adding equations (i) and (ii), we get:

`A^2(cos^2 theta + sin^2 theta)= x_0^2 + ((v_0^2)/(omega^2))`

`:. A= sqrt(x_0^2 + (v_0/omega)^2)`

Hence, the amplitude of the resulting oscillation is `sqrt(x_0^2 + (v_0/omega)^2)`

Answer 2

x = `alpha cos (omegat + theta)`

`v = (dx)/(dt) = -aomega sin(omegat = theta)`

When t = 0, x = `x_0` and `dx/dt = -v_0`

x₀ = `a cos theta`

and `-v_0 = -a omega sin theta` or a `sin theta = v_0/omega`

Squaring and adding (i) and (ii) we get

`a^2(cos^2 theta + sin^2 theta) = x_0^2 + (v_0^2)/(omega^2)` 

or `a  = sqrt(x_0^2 + v_0^2/omega^2)`