y″ = −y
Tonight's idea
Pull a mass \(m\) on a spring a distance \(y\) from rest and the spring pulls back with force \(-ky\) (Hooke's law). Newton's \(F = ma\) gives \(my'' = -ky\), where \(y''\), the second time derivative, is the acceleration. Divide by \(m\):
\(y'' = -\omega^2 y, \qquad \omega = \sqrt{k/m}.\)
Acceleration is proportional to displacement and points back to the centre. Which functions are a negative multiple of their own second derivative? \((\cos\omega t)'' = -\omega^2\cos\omega t\) and \((\sin\omega t)'' = -\omega^2\sin\omega t\). The equation is linear, so sums and multiples of solutions are solutions:
\(y = A\cos\omega t + B\sin\omega t.\)
Two constants, because a motion needs two starting facts: \(y(0) = A\) and \(y'(0) = \omega B\). Both repeat when \(\omega t\) grows by \(2\pi\), so the period is \(T = 2\pi/\omega\), whatever the amplitude: stiffer spring or lighter mass, faster oscillation.
A pendulum of length \(L\) at angle \(\theta\) obeys \(\theta'' = -(g/L)\sin\theta\). For small angles \(\sin\theta \approx \theta\), so \(\omega = \sqrt{g/L}\) and \(T = 2\pi\sqrt{L/g}\): no mass, and no amplitude for small swings, which is why pendulum clocks keep time.
Resonance. Drive the system with a force that rises and falls like a sine wave. At its own frequency \(\omega\) every push arrives in step and the swings grow and grow. At any other frequency the pushes drift in and out of step and the response stays bounded.
Watch one: \(k = 8\) N/m, \(m = 2\) kg
\(\omega = \sqrt{8/2} = 2\) rad/s, so the period is \(T = 2\pi/2 = \pi \approx 3.14\) s. Pull the mass 0.1 m and release it from rest: \(y(0) = 0.1\) gives \(A = 0.1\), and \(y'(0) = 0\) gives \(B = 0\), so \(y = 0.1\cos 2t\). Its speed is \(y' = -0.2\sin 2t\), largest, 0.2 m/s, as it passes the centre.
Instead strike it at rest with speed 0.3 m/s: \(A = 0\) and \(\omega B = 0.3\), so \(B = 0.15\) and \(y = 0.15\sin 2t\). Different start, different amplitude, same period.
Your turn
- Check that \(y = 3\cos 5t - 2\sin 5t\) satisfies \(y'' = -25y\). Find \(y(0)\), \(y'(0)\) and the period. Then write \(y\) as \(R\cos(5t + \phi)\) to find the amplitude, the largest value \(|y|\) reaches.
- A pendulum has \(L = 1.00\) m and \(g = 9.8\) m/s². Find \(\omega\) and the period. What length gives a period of exactly 2.00 s?
- With \(y = A\cos\omega t\), compute the kinetic energy \(\tfrac12 my'^2\) and the spring's stored energy \(\tfrac12 ky^2\), and show their sum is constant. What is the constant, and where is the mass when all the energy is kinetic?
- A mass hangs from a spring under gravity. Measuring \(y\) downward from the unstretched length, \(my'' = -ky + mg\). Show that \(u = y - mg/k\) satisfies \(u'' = -\omega^2 u\). What does gravity change about the motion, and what does it leave alone?
Stretch: resonance, exactly
Push at the natural frequency: \(y'' + y = \cos t\) (so \(\omega = 1\), with a unit push). Show that \(y = \tfrac12 t\sin t\) is a solution. What happens to the amplitude as \(t\) grows, and what stops it in a real spring?
Answers
- \(y'' = -75\cos 5t + 50\sin 5t = -25y\). \(y(0) = 3\), \(y'(0) = -10\), period \(2\pi/5 \approx 1.26\). Expanding \(R\cos(5t + \phi) = R\cos\phi\cos 5t - R\sin\phi\sin 5t\) and matching: \(R\cos\phi = 3\), \(R\sin\phi = 2\), so \(R = \sqrt{13} \approx 3.61\) and \(\tan\phi = \tfrac23\). Amplitude \(\sqrt{13}\).
- \(\omega = \sqrt{9.8} = 3.13\) rad/s, \(T = 2\pi/3.13 = 2.01\) s. For \(T = 2.00\) s: \(L = gT^2/(4\pi^2) = 9.8/\pi^2 = 0.993\) m, the "seconds pendulum".
- \(y' = -A\omega\sin\omega t\), so kinetic \(= \tfrac12 mA^2\omega^2\sin^2\omega t = \tfrac12 kA^2\sin^2\omega t\), using \(m\omega^2 = k\). Stored \(= \tfrac12 kA^2\cos^2\omega t\). Sum \(= \tfrac12 kA^2\), constant. All kinetic when \(\cos\omega t = 0\), that is \(y = 0\): passing the centre at top speed.
- \(u'' = y''\) since \(mg/k\) is constant, and \(-ky + mg = -k(y - mg/k) = -ku\). So \(mu'' = -ku\), \(u'' = -\omega^2u\). Gravity shifts the centre of the oscillation down by \(mg/k\), the stretch at which the spring just holds the weight. The period, \(2\pi\sqrt{m/k}\), is unchanged.
Stretch. \(y' = \tfrac12\sin t + \tfrac12 t\cos t\), and \(y'' = \tfrac12\cos t + \tfrac12\cos t - \tfrac12 t\sin t = \cos t - \tfrac12 t\sin t\). So \(y'' + y = \cos t\). The amplitude is \(\tfrac12 t\): it grows without bound, which is the right-hand panel of the figure. A real spring has friction, which removes energy at a rate that grows with the speed; the swing settles where the push's energy and the friction's losses balance, higher the less friction there is.