↓ PHYS 221.501 | More on Simple Harmonic Motion

Total mechanical energy is conserved in SHM

\(E = \frac{1}{2}mv_x^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2 = constant\)

Choose \(t’\) such that \(x(t’) = Acos(\omega t’ + \phi) = -\frac{1}{2}A\)

\(v_x(t’) = -\omega A sin(\omega t’ + \phi)) -> (v_x = -\omega Asin(\frac{2\pi}{3}) = \) oh well….

Energy diagrams for SHM

Graph these with respect to constant system energy:

\(PE = \frac{1}{2}kA^2 cos^2(\omega t + \phi)\)

\(KE = \frac{1}{2}kA^2 sin^2(\omega t + \phi)\)

If you graph the potential energy \(U\) and the total energy \(E\), \(E\) will be at a constant height, and \(U\) will be in the shape of a parabola.

The max \(U\) will be when \(cos^2(\omega t + \phi)\) is at a maximum, so it will reduce to \(\frac{1}{4}kA^2\), describing the max \(U\).

The max \(K\) will be when \(sin^2(\omega t + \phi)\) is at a maximum, so it will reduce to \(\frac{1}{4}kA^2\), describing the max \(K\), which is the same as the max \(U\).

\(x_0 = x(0) = Acos(\phi)\)

\(\omega = \sqrt{\frac{k}{m}}\)

\(x(t) = Acos(\omega t + \phi)\)

\(v_x(t) = \frac{dx}{dt} = -\omega Asin(\omega t + \phi)\)

\(T = \frac{1}{f} = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}\)

\(F = -kx\)

Energy and Momentum in SHM

📸 of a mass on a spring describing the equilibrium position, amplitude, k-ratio, and velocity.

\(Total\ energy = E_1 = \frac{1}{2}kA_1^2 = \frac{1}{2}Mv_1^2\) \(v_1 = \sqrt{\frac{k}{M}A_1}\)

If we drop putty with a mass of \(m\), momentum is conserved.

\(Mv_1 = (M + m)v_2\) \(v_2 = \frac{M}{M + m}v_1\)

\(E_2 = \frac{1}{2}(M + m)v_2^2 = \frac{1}{2}\frac{M^2}{M + m}v_1^2 = \frac{M}{M + m}(\frac{1}{2}Mv_1^2) = \frac{M}{M + m}E_1\)

\(E_2 = \frac{1}{2}kA_2^2 = \frac{M}{M + m}E_1\)

Vertical SHM

If a body oscillates from a spring, the restoring force has magnitude (kx). Therefore the vertical motion is SHM.

\(F = k\delta l = mg\)

If we have a weight on a vertically suspended spring, the net force is:

\(F_{net} = k(\delta l -x) -mg = k\delta l -kx -mg = -kx\)

Angular SHM

A coil spring exerts a restoring torque \(\tau_z = I\alpha = -\kappa\theta\) where \(\kappa\) is called the torsion.

Get a screenshot of page 441 of the book.

Remember: \(T\) describes the time to go from \(+A\) to \(+A\), not \(+A\) to \(-A\).

\(\phi = \arctan(-\frac{v_{0x}}{\omega x_0})\)

\(\tan(\phi) = -\frac{v_{0x}}{\omega x_0}\)

\(v_{0x} = -\omega A\sin(\phi)\)

\(x_0 = A\cos(\phi)\)