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 $$-A$$.

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

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

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

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