David Bauer Physics & Astronomy · UCLA

Physics 5B Chapter 16

Energy in SHM and the Pendulum

Energy in simple harmonic motion

Mechanical energy of a mass on a spring

Recall from Physics 5A that the total mechanical energy of a system is the sum of its kinetic and potential energies, $E_{\mathrm{mech}} = K + U$. For the mass-spring system, the kinetic energy of the block and the elastic potential energy stored in the spring are

\[K = \tfrac{1}{2} m v_x^2, \qquad U_{\mathrm{el}} = \tfrac{1}{2} k_s x^2.\]

Assuming no friction or external forces, the total mechanical energy

\[E_{\mathrm{mech}} = K + U_{\mathrm{el}}\]

stays constant. As the block oscillates, energy changes back and forth between kinetic and potential, but the total never changes.

Derivation: The energy of an oscillator

Since $E_{\mathrm{mech}}$ is the same at every instant, we can find it by looking at whichever instant is easiest. The simplest choice is a turning point.

At a turning point, the block is at $x = \pm A$ and is momentarily at rest, so $v_x = 0$. The kinetic energy is zero, and all of the energy is elastic potential energy:

\[K = 0, \qquad U_{\mathrm{el}} = \tfrac{1}{2} k_s A^2.\]

So the mechanical energy at the turning point, and therefore at every instant, is

\[E_{\mathrm{mech}} = K + U_{\mathrm{el}} = 0 + \tfrac{1}{2} k_s A^2.\]

Mechanical energy and amplitude

The mechanical energy of an oscillator depends only on the spring constant and the amplitude:

\[\boxed{E_{\mathrm{mech}} = \tfrac{1}{2} k_s A^2.}\]

For a given spring, doubling the amplitude quadruples the energy. In phase space, an oscillator with greater energy traces out a larger circle.

The phase determines how the total energy divides between kinetic and potential energy. The potential energy is $U_{\mathrm{el}} = \tfrac{1}{2} k_s x^2 = \tfrac{1}{2} k_s A^2\cos^2\phi$, a fraction $\cos^2\phi$ of the total. The kinetic energy, $K = E_{\mathrm{mech}} - U_{\mathrm{el}}$, is a fraction $1 - \cos^2\phi = \sin^2\phi$.

Energy bar charts

These energy bar charts show how the mechanical energy is split up at three different phases.

Three panels, each showing a small reference circle with the state point at a given phase, beside an energy bar chart with three bars: K, U sub el, and E sub mech. Kinetic energy is shown as a green hatched bar and elastic potential energy as a solid blue bar; the E sub mech bar is a stack of the two and always has the same total height. Left panel, phi equals 0, a turning point: K is zero, U sub el is the full height, and E sub mech is all potential. Middle panel, phi equals pi over 4: K and U sub el are each half the full height, and E sub mech is half kinetic and half potential. Right panel, phi equals pi over 2, passing through equilibrium: U sub el is zero, K is the full height, and E sub mech is all kinetic.

At a turning point, all of the energy is potential. As the mass passes its equilibrium position, all of it is kinetic.

Energy as a function of time

Here are the energies plotted against time for $\phi_0 = 0$. As the kinetic energy increases, the potential energy decreases by the same amount, so their sum stays fixed. Each energy form repeats twice for every cycle of the motion, because the block is at rest twice per cycle, once at each turning point.

A graph of energy versus time over one period T, with tick marks at T over 4, T over 2, 3T over 4, and T. A thick black horizontal line at height one half k sub s A squared is labeled E sub mech. A solid blue curve labeled U sub el starts at the full height at t equals 0, falls to zero at T over 4, rises to the full height at T over 2, falls to zero at 3T over 4, and returns to the full height at T. A dashed green curve labeled K is its mirror image, starting at zero and reaching the full height at T over 4 and 3T over 4. At every time the two curves add up to the constant total.

Energy and the maximum speed

At equilibrium, $U_{\mathrm{el}} = 0$, so all of the energy is kinetic. Conservation of mechanical energy gives $\tfrac{1}{2} m v_{\mathrm{max}}^2 = \tfrac{1}{2} k_s A^2$. Solving for the maximum speed, we get $v_{\mathrm{max}} = A\sqrt{k_s/m} = A\omega$, which is what we found from $v_x(t)$.

Energy conservation also gives the speed at any position, without needing the time or the phase.

Example: Speed from energy

The $0.500\ \mathrm{kg}$ block from the example in the chapter on simple harmonic motion oscillates on a spring with $k_s = 20.0\ \mathrm{N/m}$ with an amplitude of $10.0\ \mathrm{cm}$. What is its mechanical energy? How fast is it moving when it is $5.0\ \mathrm{cm}$ from equilibrium?

Solution

The mechanical energy is

\[E_{\mathrm{mech}} = \tfrac{1}{2} k_s A^2 = \tfrac{1}{2}(20.0\ \mathrm{N/m})(0.100\ \mathrm{m})^2 = 0.100\ \mathrm{J}.\]

At $x = 5.0\ \mathrm{cm}$, the elastic potential energy is

\[U_{\mathrm{el}} = \tfrac{1}{2} k_s x^2 = \tfrac{1}{2}(20.0\ \mathrm{N/m})(0.050\ \mathrm{m})^2 = 0.025\ \mathrm{J},\]

so the kinetic energy is $K = E_{\mathrm{mech}} - U_{\mathrm{el}} = 0.075\ \mathrm{J}$, and the speed is

\[v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(0.075\ \mathrm{J})}{0.500\ \mathrm{kg}}} = 0.55\ \mathrm{m/s}.\]

The simple pendulum

Describing a pendulum

A simple pendulum is a small mass $m$ hanging from a massless string or rod of length $L$ and swinging under the influence of gravity.

We describe its position by the angle $\theta$ between the string and the vertical, or by the arc length $s$ measured along the bob’s path from the equilibrium point. With $\theta$ in radians,

\[s = L\theta.\]

A simple pendulum. A string of length L hangs from a pivot on a hatched ceiling, and a round bob of mass m hangs at its end. A dashed vertical line from the pivot marks the equilibrium position. The string is pulled to the right of vertical by an angle theta, marked at the pivot. A faint dashed arc shows the circular path the bob follows as it swings. A curved arrow along this path, from the equilibrium point to the bob, is labeled s, the arc length.

Newton’s second law for the pendulum

There are two forces on the mass: tension and gravity. The tension points along the string, perpendicular to the bob’s path, so it has no component along the path. The weight has a component $mg\sin\theta$ along the path, pointing back toward equilibrium.

Taking $+s$ in the direction of increasing $\theta$, Newton’s second law along the path is

\[F_{\mathrm{net},s} = -mg\sin\theta = m a_s.\]

The pendulum bob displaced by angle theta to the right of vertical, with its force diagram. A red tension arrow labeled F sub T points from the bob up along the string toward the pivot. A red weight arrow labeled m g points straight down. A dashed red arrow labeled m g sine theta, the component of the weight along the path, points from the bob along the tangent to the path, down and to the left, back toward equilibrium. A dotted line along the tangent, labeled plus s, shows the direction of increasing theta. The angle between the weight and the extension of the string beyond the bob is also theta.

The small-angle approximation

The restoring force is proportional to $\sin\theta$, not to $\theta$, so the pendulum does not quite fit the SHM pattern. But for small angles measured in radians, $\sin\theta \approx \theta$:

$\theta$ (degrees) $\theta$ (radians) $\sin\theta$ Difference
$5^\circ$ 0.0873 0.0872 0.1%
$10^\circ$ 0.1745 0.1736 0.5%
$20^\circ$ 0.3491 0.3420 2.1%
$30^\circ$ 0.5236 0.5000 4.7%
$45^\circ$ 0.7854 0.7071 11%

The last column is how much larger $\theta$ is than $\sin\theta$. This small-angle approximation is accurate to within about 1% up to $15^\circ$, and it gets steadily worse at larger angles.

The pendulum moves in SHM

With $\sin\theta \approx \theta$, Newton’s second law becomes $m a_s = -mg\theta$. Substituting $\theta = s/L$ and canceling the mass,

\[a_s = -\frac{g}{L}\, s.\]

The acceleration is proportional to minus the displacement, which shows that this is SHM. Comparing with $a_x = -\omega^2 x$, the angular frequency of a simple pendulum is

\[\boxed{\omega = \sqrt{\frac{g}{L}}.}\]

The period $T = 2\pi\sqrt{L/g}$ depends only on the length and on $g$. The mass canceled, so a heavy bob and a light one on strings of the same length have the same period. For small angles, the period does not depend on the amplitude either.

Two different circles

The bob really does move along a physical circle of radius $L$, but only back and forth over a short arc, speeding up and slowing down. The reference circle is a circle in phase space, and its state point goes all the way around at the steady rate $\omega$.

Left, labeled real space: a pendulum hangs from a pivot, and the bob's path is a short blue arc at the bottom of a faint dashed semicircle of radius L centered on the pivot; the bob swings back and forth along this arc between two turning points. Right, labeled phase space: axes s horizontal and minus v sub s over omega vertical, with a full reference circle of radius equal to the amplitude A traced counterclockwise by a state point at a steady rate omega. Its horizontal coordinate is the arc length s of the bob.

The angular frequency $\omega = \sqrt{g/L}$ is the rate at which the phase changes, not the rate at which the string swings through the angle $\theta$.

Summary: two simple harmonic oscillators

The mass on a spring and the simple pendulum both have a restoring acceleration proportional to the displacement, so both move in simple harmonic motion:

System Equation of motion Angular frequency
Mass on a spring $a_x = -(k_s/m)\,x$ $\omega = \sqrt{k_s/m}$
Simple pendulum (small angles) $a_s = -(g/L)\,s$ $\omega = \sqrt{g/L}$

In both, $x(t) = A\cos(\omega t + \phi_0)$ (or the same with $s$ in place of $x$), with $v_{\mathrm{max}} = A\omega$ and $a_{\mathrm{max}} = A\omega^2$. The period depends on the system’s physical properties but not on the amplitude.