Oscillations

These are three nested ideas.

• Periodic motion repeats after a fixed time \(T\). The Earth going round the Sun is periodic.
• Oscillatory motion is periodic motion back and forth about an equilibrium point. A swing is oscillatory; the Earth’s orbit is periodic but not oscillatory.
• Simple harmonic motion is the special oscillation where the restoring force is proportional to displacement: \(F = -kx\). Its displacement is a single sine or cosine of time.

Every SHM is oscillatory, and every oscillation is periodic. The reverse is not true.

A function of time is SHM if it can be written as \(A\sin(\omega t+\phi)\) plus, at most, a constant (which only shifts the mean position).

• \(\sin\omega t + \cos\omega t = \sqrt2\sin(\omega t + \pi/4)\): SHM with amplitude \(\sqrt2\).
• \(\sin^2\omega t = \tfrac12 - \tfrac12\cos2\omega t\): SHM about the mean \(\tfrac12\), with angular frequency \(2\omega\) and period \(\pi/\omega\).
• \(\sin\omega t + \sin2\omega t\): periodic (period \(2\pi/\omega\)) but not SHM, because two frequencies are mixed.
• \(\sin^3\omega t\): periodic but not SHM (it contains \(\sin3\omega t\)).
• \(e^{-\omega t}\) or \(\log\omega t\): not periodic at all.

For a force law, check that \(F\) is proportional to \(-x\). A force like \(F = -kx^3\) is restoring and gives periodic motion, but it is not SHM, and its period depends on the amplitude.

If a body has potential energy \(U(x)\), equilibrium is where \(dU/dx = 0\), and it is stable at a minimum. Near that minimum, \(U\) looks like a parabola, so the force is approximately \(F = -k(x-x_0)\) with \(k = d^2U/dx^2\) at \(x_0\). This is why almost any stable system makes SHM for small displacements. Then \(\omega = \sqrt{k/m}\).

Imagine a point P moving round a circle of radius \(A\) at steady angular speed \(\omega\). Drop a perpendicular from P to a diameter. The foot of that perpendicular, N, moves back and forth along the diameter in SHM: \(x = A\cos(\omega t+\phi)\).

The angle \(\omega t+\phi\) is the phase. One full turn is \(2\pi\) and takes one period. The circle makes time questions easy: the time to go from one position to another is the angle turned by P divided by \(\omega\).

If at \(t = 0\) the body is at \(x_0\) with velocity \(v_0\), then for \(x = A\cos(\omega t+\phi)\):

1. \(x_0 = A\cos\phi\) and \(v_0 = -A\omega\sin\phi\).
2. Square and add: \(A = \sqrt{x_0^2 + (v_0/\omega)^2}\).
3. Divide: \(\tan\phi = -v_0/(\omega x_0)\). Check the signs of \(x_0\) and \(v_0\) to pick the right quadrant.

The same position with opposite velocity gives a different phase. At \(x = A/2\) moving toward the mean, \(\phi = \pi/3\); moving away from the mean, \(\phi = -\pi/3\).

Measured from the mean position, the phase angle \(\theta\) satisfies \(x = A\sin\theta\). The time taken is \(t = \theta/\omega = \theta T/2\pi\).

• Mean to \(A/2\): \(\theta = \pi/6\), \(t = T/12\).
• Mean to \(A/\sqrt2\): \(\theta = \pi/4\), \(t = T/8\).
• Mean to \(\sqrt3A/2\): \(\theta = \pi/3\), \(t = T/6\).
• \(A/2\) to \(A\): \(T/4 - T/12 = T/6\).

The body takes less time to cover the half of the path near the mean, because it moves fastest there.

Start from \(x = A\sin(\omega t+\phi)\).

1. \(v = dx/dt = A\omega\cos(\omega t+\phi)\). Using \(\cos^2 = 1-\sin^2\), \(v = \pm\omega\sqrt{A^2-x^2}\).
2. \(a = dv/dt = -A\omega^2\sin(\omega t+\phi) = -\omega^2x\).
3. Maximum speed \(v_{\max} = A\omega\) at the mean. Maximum acceleration \(a_{\max} = A\omega^2\) at the extremes.
4. Two handy combinations: \(a_{\max}/v_{\max} = \omega\) and \(v_{\max}^2/a_{\max} = A\).

• Against time: \(x\), \(v\) and \(a\) are all sinusoids of the same period. \(v\) leads \(x\) by \(\pi/2\); \(a\) leads \(v\) by \(\pi/2\), so \(a\) is opposite in phase to \(x\).
• \(v\) against \(x\): an ellipse, \(\dfrac{x^2}{A^2}+\dfrac{v^2}{A^2\omega^2} = 1\). Its horizontal semi-axis is \(A\) and its vertical semi-axis is \(A\omega\). It becomes a circle if the axes are scaled so that \(A\omega\) and \(A\) look equal, or if \(\omega = 1\).
• \(a\) against \(x\): a straight line through the origin with slope \(-\omega^2\).
• \(v^2\) against \(x^2\): a straight line with slope \(-\omega^2\) and intercept \(A^2\omega^2\).

If the speeds \(v_1\) and \(v_2\) at displacements \(x_1\) and \(x_2\) are known, write \(v_1^2 = \omega^2(A^2-x_1^2)\) and \(v_2^2 = \omega^2(A^2-x_2^2)\). Subtract to remove \(A\):

\(\omega^2 = \dfrac{v_1^2 - v_2^2}{x_2^2 - x_1^2}\), then \(A^2 = \dfrac{v_1^2x_2^2 - v_2^2x_1^2}{v_1^2 - v_2^2}\).

1. Potential energy: work done against \(F = -kx\) to stretch from 0 to \(x\) is \(U = \int_0^x kx\,dx = \tfrac12kx^2\).
2. Kinetic energy: \(K = \tfrac12mv^2 = \tfrac12m\omega^2(A^2 - x^2) = \tfrac12k(A^2 - x^2)\), using \(k = m\omega^2\).
3. Total: \(E = K + U = \tfrac12kA^2 = \tfrac12m\omega^2A^2 = 2\pi^2mf^2A^2\). It does not depend on \(x\) or \(t\).

At fixed mass, \(E\propto A^2\) and \(E\propto f^2\). Doubling both amplitude and frequency multiplies the energy by 16.

With \(x = A\sin\omega t\): \(U = E\sin^2\omega t\) and \(K = E\cos^2\omega t\). Since \(\sin^2\omega t = \tfrac12(1-\cos2\omega t)\), each energy oscillates at angular frequency \(2\omega\), that is, at twice the frequency of the motion. Its period is \(T/2\).

Averaged over a cycle, \(\langle K\rangle = \langle U\rangle = E/2\). Energy never goes negative, so it peaks twice per cycle: \(U\) at both extremes, \(K\) at each pass through the mean.

• At \(x = A/2\): \(U = E/4\), \(K = 3E/4\).
• \(K = U\) at \(x = A/\sqrt2\).
• \(K = nU\) at \(x = A/\sqrt{n+1}\). For \(K = 3U\), \(x = A/2\); for \(K = 8U\), \(x = A/3\).

Parallel. Both springs stretch by the same \(x\). Their forces add: \(F = k_1x + k_2x\), so \(k_p = k_1 + k_2\). Stiffer.

Series. The same force \(F\) passes through both springs. Their extensions add: \(x = F/k_1 + F/k_2\), so \(1/k_s = 1/k_1 + 1/k_2\), or \(k_s = k_1k_2/(k_1+k_2)\). Softer than either spring.

A block tied between two springs fixed to opposite walls is a parallel arrangement: displacing the block stretches one spring and compresses the other by the same \(x\), and both forces point back toward equilibrium.

Period rules: in series \(T_s^2 = T_1^2 + T_2^2\); in parallel \(1/T_p^2 = 1/T_1^2 + 1/T_2^2\), where \(T_1, T_2\) are the periods of the same mass on each spring alone.

For a given material and coil, \(k\propto1/L\): a shorter spring stretches less under the same force. So:

• Cut into \(n\) equal parts, each piece has stiffness \(nk\).
• Cut in the ratio \(l_1 : l_2\), the pieces have \(k_1 = k(l_1+l_2)/l_1\) and \(k_2 = k(l_1+l_2)/l_2\).

Joining the pieces back in series restores \(k\); joining them in parallel gives a much stiffer system.

• Vertical spring. At equilibrium \(kx_0 = mg\), so \(m/k = x_0/g\) and \(T = 2\pi\sqrt{x_0/g}\). The static extension alone gives the period. On a smooth incline, the period is the same \(2\pi\sqrt{m/k}\).
• Two bodies joined by a spring on a smooth floor oscillate about their centre of mass with \(T = 2\pi\sqrt{\mu/k}\), where \(\mu = m_1m_2/(m_1+m_2)\) is the reduced mass.
• Mass of the spring. A spring of mass \(m_s\) adds about \(m_s/3\) to the moving mass: \(T = 2\pi\sqrt{(m+m_s/3)/k}\).

1. Displace the bob by angle \(\theta\). Gravity \(mg\) splits into \(mg\cos\theta\) along the string (balanced with tension plus the centripetal need) and \(mg\sin\theta\) along the arc, toward the lowest point.
2. Torque about the pivot: \(\tau = -mg\ell\sin\theta\). For small \(\theta\) in radians, \(\sin\theta\approx\theta\), so \(\tau\approx-mg\ell\theta\).
3. With \(I = m\ell^2\): \(\alpha = \tau/I = -(g/\ell)\theta\). This is SHM with \(\omega^2 = g/\ell\).
4. \(T = 2\pi\sqrt{\ell/g}\). The bob’s mass cancels. The approximation is good to about 0.5% for amplitudes up to about 15°.

A seconds pendulum has \(T = 2\) s (one second per swing). Its length is \(g/\pi^2\), about 0.99 m with \(g = 9.8\) m/s².

In a non-inertial frame or with buoyancy, replace \(g\) by the effective \(g_{\rm eff}\) that the bob feels when at rest relative to the support. Then \(T = 2\pi\sqrt{\ell/g_{\rm eff}}\).

• Lift accelerating up at \(a\): \(g_{\rm eff} = g + a\), period shorter.
• Lift accelerating down at \(a\): \(g_{\rm eff} = g - a\), period longer. Free fall: \(g_{\rm eff} = 0\), no oscillation.
• Vehicle accelerating horizontally at \(a\): \(g_{\rm eff} = \sqrt{g^2 + a^2}\); the string hangs tilted at \(\tan^{-1}(a/g)\).
• Bob of density \(\rho\) swinging in a non-viscous liquid of density \(\sigma\): buoyancy gives \(g_{\rm eff} = g(1 - \sigma/\rho)\).
• Inside an orbiting satellite: \(g_{\rm eff} = 0\).

• \(\ell\) is measured from the pivot to the bob’s centre of mass. A hollow bob filled with water that slowly drains: the centre of mass first drops (period rises), then returns to the centre once empty (period falls back).
• Small length changes: \(\Delta T/T = \tfrac12\,\Delta\ell/\ell\). A metal pendulum lengthens when hot, so the clock runs slow in summer.
• For a length comparable to the Earth’s radius, \(T = 2\pi\sqrt{\dfrac{1}{g(1/\ell + 1/R)}}\). As \(\ell\to\infty\), \(T\to2\pi\sqrt{R/g}\approx84\) min, the same as a satellite skimming the surface.

1. A body of mass \(m\) with uniform cross-section \(A\) floats at rest with depth \(h\) submerged: \(mg = \sigma Ahg\).
2. Push it down by an extra \(x\). The buoyant force becomes \(\sigma A(h+x)g\), so the net upward force is \(\sigma Agx\).
3. Net force \(F = -\sigma Ag\,x\), which is SHM with \(k = \sigma Ag\).
4. \(T = 2\pi\sqrt{m/(\sigma Ag)}\). Substituting \(m = \sigma Ah\): \(T = 2\pi\sqrt{h/g}\).

If the body has height \(L\) and density \(\rho\), then \(h = L\rho/\sigma\) and \(T = 2\pi\sqrt{L\rho/(\sigma g)}\).

1. Inside a uniform Earth, at distance \(r\) from the centre, \(g(r) = gr/R\), directed to the centre.
2. Along a diametral tunnel, \(r = x\), so \(F = -(mg/R)x\). This is SHM with \(\omega = \sqrt{g/R}\).
3. \(T = 2\pi\sqrt{R/g}\approx84\) min with \(R = 6.4\times10^6\) m and \(g = 10\) m/s².
4. Speed at the centre \(= A\omega = R\sqrt{g/R} = \sqrt{gR}\approx8\) km/s, the same as the speed of a satellite skimming the surface.

For a straight tunnel along a chord (not through the centre), only the component of gravity along the tunnel acts. It is \(-(mg/R)\,s\), where \(s\) is the distance from the tunnel’s midpoint. So the period is the same 84 minutes for every chord.

To show any system does SHM and find its period:

1. Find the equilibrium position.
2. Displace by a small \(x\) and write the net restoring force (or torque).
3. Simplify for small \(x\) to the form \(F = -kx\) (or \(\tau = -C\theta\)).
4. Write \(T = 2\pi\sqrt{m/k}\) (or \(2\pi\sqrt{I/C}\)), with \(m\) the total moving mass.

A small ball rolling without slipping at the bottom of a bowl, a liquid in a U-tube and a pendulum all fit this pattern.

With a resistive force \(-bv\), the solution for light damping is

\(x = A_0e^{-bt/2m}\cos(\omega' t + \phi)\), with \(\omega' = \sqrt{\dfrac km - \dfrac{b^2}{4m^2}}\).

• The amplitude decays exponentially: \(A(t) = A_0e^{-bt/2m}\). It falls by the same fraction in each equal time interval.
• Energy goes as amplitude squared: \(E(t) = E_0e^{-bt/m}\), so energy decays twice as fast (in rate constant).
• The time for the amplitude to halve is \(t_{1/2} = 2m\ln2/b\).
• Damping slightly lowers the frequency (\(\omega' < \omega_0\)). Very heavy damping stops oscillation altogether.

A periodic force \(F_0\cos\omega_dt\) drives the oscillator. After the starting transients die away, it oscillates at the driving frequency \(\omega_d\), with steady amplitude

\(A = \dfrac{F_0/m}{\sqrt{(\omega_0^2 - \omega_d^2)^2 + (b\omega_d/m)^2}}\).

• If \(\omega_d\) is far from \(\omega_0\), the amplitude is small.
• Near \(\omega_d = \omega_0\), the first term vanishes and the amplitude is limited only by damping: \(A\approx F_0/(b\omega_0)\). This is resonance.
• Light damping gives a tall, narrow peak; heavy damping gives a low, broad one.

• Soldiers break step when crossing a bridge so that their marching does not drive the bridge at its natural frequency.
• A radio is tuned by changing its circuit’s natural frequency to match the station.
• A singer’s note at the natural frequency of a wine glass can make it vibrate strongly enough to crack.
• Pushing a swing at the right moment each cycle builds a large amplitude with small pushes.

Two SHMs of the same frequency along the same line, \(x_1 = A_1\sin\omega t\) and \(x_2 = A_2\sin(\omega t+\delta)\), add like vectors (phasors) of lengths \(A_1, A_2\) with angle \(\delta\) between them:

\(A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\delta}\), \(\tan\phi = \dfrac{A_2\sin\delta}{A_1 + A_2\cos\delta}\).

• \(\delta = 0\): \(A = A_1 + A_2\). \(\delta = \pi\): \(A = |A_1 - A_2|\). \(\delta = \pi/2\): \(A = \sqrt{A_1^2 + A_2^2}\).
• Equal amplitudes \(a\) with phase difference \(\delta\): \(A = 2a\cos(\delta/2)\). At \(\delta = 2\pi/3\), \(A = a\).
• Different frequencies do not give SHM.

For \(x = a\sin\omega t\) and \(y = b\sin(\omega t + \delta)\) of the same frequency:

• \(\delta = 0\) or \(\pi\): a straight line \(y = \pm(b/a)x\). The motion is SHM along a slanted line with amplitude \(\sqrt{a^2+b^2}\).
• \(\delta = \pi/2\): an ellipse \(x^2/a^2 + y^2/b^2 = 1\); a circle if \(a = b\) (uniform circular motion).
• Other \(\delta\): a tilted ellipse.

1. Push the liquid down by \(x\) on one side; it rises by \(x\) on the other. The level difference is \(2x\).
2. The unbalanced pressure \(\rho g(2x)\) acts on the cross-section \(A\): restoring force \(F = -2\rho gAx\).
3. The whole liquid column moves: mass \(m = \rho AL\), where \(L\) is the total length of liquid.
4. \(a = F/m = -(2g/L)x\), so \(\omega^2 = 2g/L\) and \(T = 2\pi\sqrt{L/2g}\). Density and area cancel.
5. If \(h\) is the height of liquid in each arm at rest (with negligible bottom length), \(L = 2h\) and \(T = 2\pi\sqrt{h/g}\).

Floating bodies and a tunnel through the Earth are treated in “Other SHM systems”.