At a village mela a child rides a big wooden swing. At each end of the ride the swing stops for an instant, high in the air, and the child squeals. At the bottom she rushes past fastest, close to the ground. The height and the speed trade places again and again, but the ride keeps its size for a long time. Something is being passed back and forth without being lost.
Two stores of energy
A block oscillates on a smooth floor on a spring of constant \(k\), with amplitude \(A\). Its energy sits in two stores. The block has kinetic energy \(K = \tfrac12 mv^2\) because it moves. The stretched or compressed spring has potential energy \(U = \tfrac12 kx^2\), as you found in the chapter Potential Energy and Conservation of Energy. Both stores change at every instant.
The chapter Kinematics of SHM gave the speed at any position, \(v^2 = \omega^2(A^2 - x^2)\). Put it into \(K\), and use \(m\omega^2 = k\). The kinetic energy is \(\tfrac12 m\omega^2(A^2 - x^2) = \tfrac12 k(A^2 - x^2)\). It is largest at the mean position, where it is \(\tfrac12 kA^2\), and zero at the extremes \(x = \pm A\), where the block stops.
The potential energy does the opposite. \(U = \tfrac12 kx^2\) is zero at the mean position, where the spring has its natural length. It is largest at the extremes, \(\tfrac12 kA^2\), where the spring is stretched or compressed most. As the block moves out, energy passes from the block's motion into the spring. As it moves back, the spring hands the energy back.
In SHM, where is the kinetic energy of the oscillating body greatest?
The energies against time
Write the motion as \(x = A\sin(\omega t + \varphi)\), so that \(v = A\omega\cos(\omega t + \varphi)\). Then the two energies become functions of time. The potential energy is \(\tfrac12 kA^2\sin^2(\omega t + \varphi)\). The kinetic energy is \(\tfrac12 mA^2\omega^2\cos^2(\omega t + \varphi)\), which is \(\tfrac12 kA^2\cos^2(\omega t + \varphi)\). One is a sine squared, the other a cosine squared.
A sum that never changes
Add the two stores. The \(x^2\) terms cancel, and what remains does not depend on the position at all. In the time form, \(\sin^2 + \cos^2 = 1\) does the same job. The total mechanical energy of the oscillator is constant, as it must be: the spring force is conservative and nothing else does work.
\[ E = K + U = \tfrac12 m\omega^2(A^2 - x^2) + \tfrac12 kx^2 \]
Why. the kinetic energy from \(v^2 = \omega^2(A^2 - x^2)\) plus the potential energy of the spring.
\[ E = \tfrac12 k(A^2 - x^2) + \tfrac12 kx^2 \]
Why. \(\omega^2 = k/m\), so \(m\omega^2 = k\).
\[ E = \tfrac12 kA^2 - \tfrac12 kx^2 + \tfrac12 kx^2 \]
Why. open the bracket.
\[ E = \tfrac12 kA^2 = \tfrac12 m\omega^2A^2 \]
Read it. the \(x^2\) terms cancel; the total is the same at every position.
In SHM with no friction, the total mechanical energy is \(E = \tfrac12 kA^2 = \tfrac12 m\omega^2A^2\) at every position and every instant. It depends only on the force constant and the square of the amplitude.
Fig 7.2 turns the energy law into geometry. The potential energy parabola opens upward from the mean; the kinetic energy parabola opens downward from the top of the line. At every \(x\) the two heights add up to the line. Where the parabola \(U\) meets the line, at \(x = \pm 0.2\,\text{m}\), the kinetic energy is zero. These are the turning points of the chapter Potential Energy and Conservation of Energy. The block can never go beyond them.
The amplitude of an oscillator in SHM is doubled, with the same spring and mass. What happens to its total energy?
At some position a body in SHM has potential energy \(U = \tfrac12 kA^2\). What is its kinetic energy there?
A \(0.5\,\text{kg}\) block oscillates on a spring of \(200\,\text{N/m}\) with an amplitude of \(0.1\,\text{m}\). Find its total energy, and its kinetic energy, potential energy and speed at \(x = 6\,\text{cm}\).
The total energy is fixed by the amplitude. At any \(x\) the potential energy is \(\tfrac12 kx^2\), and the rest is kinetic.
\(E = \tfrac12 kA^2\), \(U = \tfrac12 kx^2\), \(K = E - U = \tfrac12 mv^2\).
\(E = \tfrac12 \times 200 \times 0.01 = 1.0\,\text{J}\). \(U = \tfrac12 \times 200 \times 0.0036 = 0.36\,\text{J}\). \(K = 0.64\,\text{J}\), so \(v = \sqrt{2 \times 0.64/0.5} = 1.6\,\text{m/s}\).
At \(6\,\text{cm}\) the energy is \(64\,\%\) kinetic and \(36\,\%\) potential, and the block moves at \(1.6\,\text{m/s}\).
Where the two stores are equal
At the mean the energy is all kinetic; at the extremes it is all potential. Where are the two equal? Set \(\tfrac12 kx^2 = \tfrac12 k(A^2 - x^2)\). Then \(2x^2 = A^2\) and \(x = \pm A/\sqrt{2}\), about \(0.71A\). Each store then holds half the total, \(\tfrac14 kA^2\). The point is not halfway out, because the potential energy grows as the square of \(x\). At \(x = A/2\) the spring holds only a quarter of the energy.
At which position in SHM is the kinetic energy equal to the potential energy?
In time, the two energies trade places twice every period. \(K\) goes as \(\cos^2(\omega t + \varphi)\), which repeats every half period, so the kinetic energy rises and falls with period \(T/2\). The same is true of \(U\). Starting from the mean, the body reaches \(A/\sqrt{2}\), where \(K = U\), after one eighth of a period, since \(\omega t = \pi/4\) there. Over a whole period, each store holds on average half the total, \(\tfrac14 kA^2\).
A \(1\,\text{kg}\) body moves in SHM with angular frequency \(4\,\text{rad/s}\) and amplitude \(0.5\,\text{m}\). What is its total energy, in joules?
The energy method
So far the force came first and the energy followed. Now go the other way. The chapter Potential Energy and Conservation of Energy showed where a particle rests in stable equilibrium: at a minimum of its potential energy \(U(x)\). Every smooth curve, looked at closely near its lowest point \(x_0\), looks like a parabola. The shape of that parabola is set by the curvature, the second derivative \(U''(x_0)\).
\[ U(x) \approx U(x_0) + U'(x_0)(x - x_0) + \tfrac12 U''(x_0)(x - x_0)^2 \]
Why. the first terms of the Taylor expansion of \(U\) about \(x_0\), good for small \(x - x_0\).
\[ U(x) \approx U(x_0) + \tfrac12 U''(x_0)(x - x_0)^2 \]
Why. at a minimum the slope is zero, \(U'(x_0) = 0\).
\[ F = -\dfrac{dU}{dx} \approx -U''(x_0)\,(x - x_0) \]
Why. the force is minus the slope of \(U\), from the chapter Potential Energy and Conservation of Energy.
\[ k_{\text{eff}} = U''(x_0), \qquad \omega^2 = \dfrac{U''(x_0)}{m} \]
Read it. a linear restoring force about \(x_0\); the two steps give \(\omega\).
This is the energy method. It explains why SHM is everywhere. Any system nudged a little from a stable equilibrium sits near the bottom of a parabola, so it oscillates in SHM, whatever the forces are. The method also reaches cases where the force is awkward to write down, because an energy is often easier to find than a force.
A particle of mass \(0.2\,\text{kg}\) moves along the \(x\) axis with potential energy \(U = 20/x + 5x\), for \(x > 0\), with \(U\) in joules and \(x\) in metres. Find its position of stable equilibrium and the period of small oscillations about it.
\(U' = -20/x^2 + 5\), which is zero at \(x_0 = 2\,\text{m}\). There \(U = 10 + 10 = 20\,\text{J}\).
\(U'' = 40/x^3\), which is \(5\,\text{N/m}\) at \(x_0 = 2\,\text{m}\). It is positive, so the equilibrium is stable.
\(\omega = \sqrt{U''/m} = \sqrt{5/0.2} = 5\,\text{rad/s}\), so \(T = 2\pi/5 = 1.26\,\text{s}\).
The energy method gives the U-tube of lesson 6 in two lines. The liquid of mass \(\rho AL\) moves with speed \(v\), so \(K = \tfrac12\rho ALv^2\). Raising a column of height \(x\) and cross-section \(A\) by \(x\) stores \(U = \rho Agx^2\). Written as \(\tfrac12 k_{\text{eff}}x^2\), that is \(k_{\text{eff}} = 2\rho Ag\), and \(\omega^2 = 2\rho Ag/\rho AL = 2g/L\), as before.
A particle of mass \(1.5\,\text{kg}\) has potential energy \(U = 3x^2 - 12x + 20\), with \(U\) in joules and \(x\) in metres. What is the angular frequency of its small oscillations about the minimum, in rad/s?
The swing at the mela is an energy account. At each end of the ride the swing stops, and its energy is all potential, stored in its height. At the bottom the height is least and the energy is all kinetic, so the child moves fastest. The total stays nearly constant, so the ride keeps its size. Halfway up the arc is not where the two are equal: that happens at \(A/\sqrt{2}\), about seven tenths of the way out.
- Putting the equal-energy point halfway out.\(U = \tfrac12 kx^2\) grows as \(x^2\), so \(K = U\) at \(x = \pm A/\sqrt{2}\). At \(A/2\) the energy is only a quarter potential.
- Thinking that the energy doubles with the amplitude.\(E = \tfrac12 kA^2\): twice the amplitude is four times the energy.
- Taking the period of \(K\) or \(U\) to be \(T\).\(K\) and \(U\) each repeat every half period, \(T/2\), because they go as the square of a sine or cosine.