More about energy in SHM IB DP Physics Study Notes - 2025 Syllabus

The defining equation of SHM: a = -ω²x

• Note that the shadow is the x-coordinate of the ball.
• 
• Thus the equation of the shadow’s displacement is
• x = x₀ cos θ.
• Since ω = θ/t we can write θ = ωt.
• Therefore the equation of the shadow’s x-coordinate i
• x = x₀ cos ωt.
• If we know ω, and if we know t, we can then calculate x.
• Similarly, if the object is starting at a displacement of 0, these equations can be applied to it.:
• 
• This set of equations is derived by observing the shadow from a light at the for the picture at the right, beginning as shown:

• The blue oscillation is starting $\frac{3\pi}{2}$ behind the red.
• The standard curve has the displacement equation:
• $x = x₀ sin ωt$
• To horizontally shift this, we have to add the phase difference to the ωt term. For the image above it would be:
• $x = x₀ sin (ωt + \frac{3\pi}{2})$
• Generally:
• $x = x₀ sin (ωt + ϕ) $

From : $x = x_{0} \cos \omega t$, $v = -x_{0} \omega \sin \omega t$ and $v = x_{0} \omega$

Begin by squaring each equation from :

$x^{2} = x_{0}^{2} \cos^{2} \omega t$,

$v^{2} = (-x_{0} \omega \sin \omega t)^{2} = x_{0}^{2} \omega^{2} \sin^{2} \omega t$.

Now $\sin^{2} \omega t + \cos^{2} \omega t = 1$ yields $\sin^{2} \omega t = 1 – \cos^{2} \omega t$

so that $v^{2} = x_{0}^{2} \omega^{2}(1 – \cos^{2} \omega t)$ or

$v^{2} = \omega^{2}(x_{0}^{2} – x_{0}^{2} \cos^{2} \omega t)$.

Then $v^{2} = \omega^{2}(x_{0}^{2} – x^{2})$, which becomes

$v = \pm \omega \sqrt{x_{0}^{2} – x^{2}}$