Suppose a block held on a frictionless plane attached to a spring(upstretched) is pulled to a distance x₀ 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 x₀ from mean position

or x = x₀ at t = 0

so putting in equation

x = a cos (ωt+θ)

we have

x₀ = a cosθ (eq -1)

now we have initial condition that velocity of particle is initially is v₀ directed towards mean position in negative x direction

or v = v₀ at t = 0

so putting in equation

v = -aω sin (ωt+θ)

we get

v₀ = -aω sinθ

or we can rewrite it as

-v₀/ω = a sin θ (eq -2)

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

x₀²+ (-v₀/ω)² = (a cos θ)² + (a sin θ)²

Solving further

x₀² + v₀²/ω² = a²cos²θ + a²sin²θ

or

a²(cos²θ + sin²θ) = x₀² + v₀²/ω²

we know the identity cos²θ + sin²θ = 1

so we have

a² = x₀² + v₀²/ω²

or the amplitude is

so the amplitude of oscillation of the block is