More about energy in SHM IB DP Physics Study Notes - 2025 Syllabus
More about energy in SHM IB DP Physics Study Notes
More about energy in SHM IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
- That a particle undergoing simple harmonic motion can be described using the phase angle \( \phi \).
- That problems can be solved using the equations for simple harmonic motion as given by:
- \(x = x_0 \sin(\omega t + \phi)\)
- \(v = \omega x_0 \cos(\omega t + \phi)\)
- \(v = \pm \omega \sqrt{x_0^2 – x^2}\)
- \(E_T = \frac{1}{2} m \omega^2 x_0^2\)
- \(E_p = \frac{1}{2} m \omega^2 x^2\)
Standard level and higher level: 3 hours
Additional higher level: 4 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
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
$v = \pm \omega \sqrt{x_{0}^{2} – x^{2}}$
Energy changes during SHM
Recall the relation between v and x that we derived in the last section: $v = \pm \omega \sqrt{x_{0}^{2} – x^{2}}$.
Then $v^{2} = \omega^{2}(x_{0}^{2} – x^{2})$
$½mv^{2} = ½m\omega^{2}(x_{0}^{2} – x^{2})$
$E_{K} = ½m\omega^{2}(x_{0}^{2} – x^{2})$.
Recall that $v_{max} = v_{0} = \omega x_{0}$ so that we have
$E_{K,max} = ½mv_{max}^{2} = ½m\omega^{2}x_{0}^{2}$.
\(\text{Since } E_P = 0 \text{ when } E_K = E_{K,\text{max}}, \text{ we have:}\)
\(E_K + E_P = E_T\)
\(E_{K,\text{max}} + 0 = E_T\)
\(\text{From } E_K = \frac{1}{2} m \omega^2 (x_0^2 – x^2), \text{ we get:}\)
\(E_K = \frac{1}{2} m \omega^2 x_0^2 – \frac{1}{2} m \omega^2 x^2\)
\(E_K = E_T – \frac{1}{2} m \omega^2 x^2\)
\(E_T = E_K + \frac{1}{2} m \omega^2 x^2\)
