C.1 Simple harmonic motion

Solution outline

1. $F = -kx \Rightarrow ma = -kx$
2. $a = -\dfrac{k}{m}x$
3. $\text{Comparing with } a = -\omega^{2}x \text{ gives } \omega^{2} = k/m.$
4. $\text{Acceleration is directly proportional to displacement and always directed towards the equilibrium position, so the motion is SHM.}$

Solution outline

1. $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{6.0} \approx 1.05\ \mathrm{s}$
2. $f = \dfrac{1}{T} \approx \dfrac{1}{1.05} \approx 0.95\ \mathrm{Hz}$

Solution outline

1. $x(t) = 0.080\cos(5.0t)$
2. $v(t) = \dfrac{dx}{dt} = -0.080\times5.0\sin(5.0t) = -0.40\sin(5.0t)\ \mathrm{m\,s^{-1}}$
3. $a(t) = \dfrac{dv}{dt} = -0.40\times5.0\cos(5.0t) = -2.0\cos(5.0t)\ \mathrm{m\,s^{-2}}$
4. $x(0.30) = 0.080\cos(1.5) \approx 0.0056\ \mathrm{m}$
5. $v(0.30) = -0.40\sin(1.5) \approx -0.40\times0.997 = -0.40\ \mathrm{m\,s^{-1}}$
6. $a(0.30) = -2.0\cos(1.5) \approx -2.0\times0.070 = -0.14\ \mathrm{m\,s^{-2}}$

Solution outline

1. $T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.40}{25}}$
2. $T = 2\pi\sqrt{0.016} \approx 2\pi\times0.126 = 0.79\ \mathrm{s}$
3. $\omega = \dfrac{2\pi}{T} \approx \dfrac{2\pi}{0.79} \approx 8.0\ \mathrm{rad\,s^{-1}}$

Solution outline

1. $T = 2\pi\sqrt{\dfrac{L}{g}} = 2\pi\sqrt{\dfrac{0.90}{9.8}}$
2. $T \approx 2\pi\sqrt{0.092} \approx 2\pi\times0.303 = 1.90\ \mathrm{s}$
3. $\text{This expression assumes small angular displacements so that } \sin\theta \approx \theta \text{ in radians.}$

Solution outline

1. $E_{\text{tot}} = \tfrac{1}{2}kA^{2} \text{ is constant throughout the motion.}$
2. $E_{\text{p}} = \tfrac{1}{2}kx^{2}$
3. $E_{\text{k}} = E_{\text{tot}} - E_{\text{p}} = \tfrac{1}{2}k(A^{2} - x^{2})$
4. $\text{At the extremes } (x = \pm A) \text{ all energy is potential and the speed is zero.}$
5. $\text{At the equilibrium position } (x = 0) \text{ all energy is kinetic and the speed is maximum.}$

Solution outline

1. $A = 0.050\ \mathrm{m},\ \omega = 10\ \mathrm{rad\,s^{-1}},\ \phi = \tfrac{\pi}{3}.$
2. $Initial displacement: x(0) = 0.050\cos\left(\tfrac{\pi}{3}\right) = 0.050\times0.5 = 0.025\ \mathrm{m}.$
3. $\text{The phase angle } \phi \text{ shifts the cosine curve horizontally; it specifies where in the cycle the oscillator is at } t = 0.$
4. $\text{Here the motion starts one-sixth of a full cycle } (60^{\circ}) \text{ ahead of a cosine that would start at maximum displacement.}$