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A Mass Attached to a Spring is Free to Oscillate, with Angular Velocity ω, in a Horizontal Plane Without Friction Or Dampinng Determine the Amplitude of the Resulting Oscillations in Terms of the Parameters ω - CBSE (Science) Class 11 - Physics

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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 = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation acos (ωt) and note that the initial velocity is negative.]

Solution 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)`

Solution 2

x = `alpha cos (omegat + theta)`

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

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

x= `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)`

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Solution A Mass Attached to a Spring is Free to Oscillate, with Angular Velocity ω, in a Horizontal Plane Without Friction Or Dampinng Determine the Amplitude of the Resulting Oscillations in Terms of the Parameters ω Concept: Some Systems Executing Simple Harmonic Motion.
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