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 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 ω, x_{0} and v_{0}. [Hint: Start with the equation

x = a cos (ωt+θ) and note that the initial velocity is negative.]

Suppose a block held on a frictionless plane attached to a spring(upstretched) is pulled to a distance x_{0} and then pushed towards the centre that is the mean position or original position with velocity v_{0,} now the block will start to oscillate and undergo Simple Harmonic Motion with an angular frequency ω, the situation at beginning of time t = 0 has been

**shown in the figure**

Let the equation of Simple harmonic motion of the block be

x = a cos (ωt+θ)

where x is denotes the position of particle or displacement from mean position at any instant of time t, ω is the angular frequency of the Simple harmonic motion of the block, θ is the initial phase of the block a is the amplitude or the maximum displacement of block from the mean position

we have to find the amplitude a

differentiating the equation resenting position of particle with respect to time to find velocity of particle at any instant

We know where v is the velocity of the particle

And differentiation where a, b and c are constants

So differentiating we get velocity of particle

v = -aω sin (ωt+θ)

now we have initial condition that initially at t = 0, particle is at a distance x_{0} from mean position

or x = x_{0} at t = 0

so putting in equation

x = a cos (ωt+θ)

we have

x_{0} = a cosθ (eq -1)

now we have initial condition that velocity of particle is initially is v_{0} directed towards mean position in negative x direction

or v = v_{0} at t = 0

so putting in equation

v = -aω sin (ωt+θ)

we get

v_{0} = -aω sinθ

or we can rewrite it as

-v_{0}/ω = a sin θ (eq -2)

Squaring and adding (eq -1) and (eq -2) we get

x_{0}^{2}+ (-v_{0}/ω)^{2} = (a cos θ)^{2} + (a sin θ)^{2}

Solving further

x_{0}^{2} + v_{0}^{2}/ω^{2} = a^{2}cos^{2}θ + a^{2}sin^{2}θ

or

a^{2}(cos^{2}θ + sin^{2}θ) = x_{0}^{2} + v_{0}^{2}/ω^{2}

we know the identity cos^{2}θ + sin^{2}θ = 1

so we have

a^{2} = x_{0}^{2} + v_{0}^{2}/ω^{2}

or the amplitude is

so the amplitude of oscillation of the block is

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