Energy in Oscillators

The question a stopwatch cannot answer

You have spent the last lesson on pendulums, finding the period of a swing and writing the angle as a cosine in time. That machinery answers "where is it at 2.4 seconds?" But most oscillator problems never mention time at all. They ask how fast the bob is moving at the bottom, or how far a block gets before it stops, or what happens to the swing when you drop a lump of putty on it. Solving those with $x(t) = A\cos\omega t$ means finding the phase, differentiating, and hoping the algebra closes.

There is a shortcut, and it is the single most useful tool in this unit. An oscillator has a fixed budget of energy. It spends that budget on stretching the spring and on moving the mass, and the split between the two changes constantly while the total sits still. Track the budget instead of the clock and most of these questions collapse to one line of algebra.

A constant total, traded back and forth

Take a block of mass $m$ on a horizontal spring of stiffness $k$, and let $x$ be the displacement from the relaxed position. Two things hold energy here. The spring holds potential energy $U = \tfrac{1}{2}kx^2$: that is the work you do stretching it, since the force $kx$ grows from zero and averages $\tfrac12 kx$ over a pull of length $x$. The block holds kinetic energy $K = \tfrac{1}{2}mv^2$.

Nothing else takes a cut. Gravity and the normal force are both perpendicular to the motion, so neither does work. With no friction, the sum is conserved:

$E = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \text{constant}$

That constant is worth naming in a useful way. Evaluate it at the turning point, where the block is momentarily at rest at $x = A$, the amplitude. There $v = 0$, so every joule sits in the spring:

$E = \tfrac{1}{2}kA^2$

Now evaluate the same constant at the centre, where $x = 0$ and the spring holds nothing. Every joule is kinetic, and the block is moving at its fastest:

The $\omega = \sqrt{k/m}$ is the same angular frequency from the last two lessons, which is a decent sign the bookkeeping is right. Notice what the energy statement does not contain: $t$. It connects a position to a speed directly, which is exactly the connection those problems ask for.

Reading a speed off the energy line

Set the total at a general position equal to the total at the turning point:

$\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2$

Multiply through by $2/m$ and use $\omega^2 = k/m$:

Read that as a sentence. The speed is largest at $x=0$, falls to zero at $x = \pm A$, and depends on $x$ only through $x^2$, so the block moves equally fast on the way out and on the way back. The mass never appears except inside $\omega$.

Two ratio facts follow immediately, and problems love them because they need no numbers at all. The potential share is $U/E = (\tfrac12 kx^2)/(\tfrac12 kA^2) = x^2/A^2$, so the kinetic share is $K/E = 1 - x^2/A^2$. At $x = 0.60A$ the kinetic fraction is $1 - 0.36 = 0.64$. And $K = U$ means each holds half, so $\tfrac12 kx^2 = \tfrac12\left(\tfrac12 kA^2\right)$, giving $x^2 = A^2/2$ and

$x = \frac{A}{\sqrt 2} \approx 0.707A$

Example 1. A 0.25 kg puck on a spring of stiffness 100 N/m swings with amplitude 8.0 cm. How fast is it moving 4.0 cm from equilibrium?

Start with $\omega$, because $v = \omega\sqrt{A^2-x^2}$ needs it: $\omega = \sqrt{100/0.25} = \sqrt{400} = 20$ rad/s. Then $A^2 - x^2 = (0.080)^2 - (0.040)^2 = 0.0064 - 0.0016 = 0.0048$ m$^2$, and $\sqrt{0.0048} = 0.0693$ m. So $v = 20 \times 0.0693 = 1.39$ m/s.

Check it against the maximum, $v_{\max} = \omega A = 20 \times 0.080 = 1.6$ m/s. Halfway out in position, the puck is still at 87% of top speed. That is the parabola doing its work: near the centre, the spring holds almost nothing.

Vertical springs and pendulums store it somewhere else

Hang the same block from the spring and gravity now does work as it moves. The fix is short. Let $y$ be the displacement measured from the hanging equilibrium, where the stretch is $y_0 = mg/k$. Add the spring and gravity terms and expand:

The cross terms cancel because $ky_0 = mg$, and what is left is $\tfrac12 ky^2$ plus a constant that never changes. So a vertical spring behaves exactly like a horizontal one, provided you measure from the new equilibrium and forget gravity entirely. A 1.5 kg mass on a 45 N/m spring pulled 8.0 cm below its hanging position reaches $v_{\max} = 0.080\sqrt{45/1.5} = 0.080 \times 5.477 = 0.438$ m/s, with $g$ appearing nowhere.

A pendulum keeps its potential energy in height instead. A bob on a string of length $L$ at angle $\theta$ from vertical sits a height $h = L(1-\cos\theta)$ above the bottom of its swing, so $U = mgL(1-\cos\theta)$. Released from rest at $\theta_0$, energy conservation to the lowest point gives $\tfrac12 mv^2 = mgL(1-\cos\theta_0)$, so

$v = \sqrt{2gL\,(1-\cos\theta_0)}$

This one is exact at any angle, small or not. For a 1.20 m string released at $15^\circ$: $1-\cos 15^\circ = 1 - 0.9659 = 0.0341$, so $v = \sqrt{2(9.8)(1.20)(0.0341)} = \sqrt{0.8014} = 0.895$ m/s.

Example 2. A bob on a 1.00 m string is released from rest at $60^\circ$ from vertical. At the lowest point, what is the ratio of string tension to weight?

Energy alone will not answer this, because tension does no work and so never appears in the energy equation. It gives you the speed; Newton's second law turns the speed into a force. Do them in that order.

Speed first: $v^2 = 2gL(1-\cos 60^\circ) = 2gL(1-0.5) = gL$.

Now the bottom of the swing. The bob moves on a circle of radius $L$, so its acceleration there points straight up, towards the pivot, with magnitude $v^2/L$. Two vertical forces act: tension $T$ up and weight $mg$ down.

$T - mg = \frac{mv^2}{L} = \frac{m(gL)}{L} = mg$

So $T = 2mg$ and the ratio is exactly 2. A cleaner way to hold it: at the bottom, $T/mg = 3 - 2\cos\theta_0$, which you can get by leaving $\theta_0$ general. At $60^\circ$ that is $3-1 = 2$.

Averages over a cycle

Put $x = A\cos\omega t$ into both energies, with $v = -A\omega\sin\omega t$ and $m\omega^2 = k$:

Over a full cycle the average of $\cos^2$ and of $\sin^2$ is $\tfrac12$ each, since the two averages must be equal and must add to 1. So

$\langle U\rangle = \langle K\rangle = \tfrac{1}{4}kA^2 = \tfrac{1}{2}E$

On average, the energy splits down the middle. For a 0.40 kg mass on a 90 N/m spring with $A = 5.0$ cm, the mean kinetic energy is $\tfrac14 (90)(0.050)^2 = \tfrac14 (90)(0.0025) = 0.0563$ J.

Both $\cos^2$ and $\sin^2$ run through two full periods for every one period of the motion, so the energies slosh at $2\omega$ while the block oscillates at $\omega$. That factor of two is behind a lot of otherwise mysterious answers.

When the mass changes mid-oscillation

A popular problem type drops a lump of putty on a moving block, or lifts a rider off it, and asks for the new amplitude. Energy is not conserved in a sticking collision, so you cannot start there. What you can do is ask, at the instant of the change, which of $x$ and $v$ survives it, then rebuild the energy from whichever survived.

At the centre, the block is moving and $x = 0$. A vertical drop of putty adds no horizontal momentum, so horizontal momentum is conserved: $mv = (m+M)v'$. The new state is "speed $v'$ at $x=0$", so the whole new energy is kinetic and $\tfrac12(m+M)v'^2 = \tfrac12 kA'^2$.

At the turning point, $v=0$ and the spring is stretched by $A$. Adding or removing mass there changes no stored energy at all, so the amplitude is unchanged. Only the speed later on changes, through the new $\omega$.

Example 3. A 0.60 kg block oscillates on a $k = 240$ N/m spring with amplitude 10.0 cm. A 0.30 kg lump of putty is dropped onto it and sticks. Find the new amplitude if the drop happens at the centre, and the new maximum speed if it happens at the turning point instead.

Dropped at the centre. The speed there before the drop is $v = \omega A = \sqrt{240/0.60}\times 0.100 = \sqrt{400}\times 0.100 = 20 \times 0.100 = 2.00$ m/s. Momentum: $(0.60)(2.00) = (0.90)v'$, so $v' = 1.20/0.90 = 1.333$ m/s. That 0.9 kg moving at 1.333 m/s carries $\tfrac12(0.90)(1.333)^2 = 0.45 \times 1.778 = 0.800$ J. Setting $\tfrac12(240)A'^2 = 0.800$ gives $A'^2 = 1.600/240 = 0.006667$ m$^2$ and $A' = 0.0817$ m. The amplitude shrank, because the collision threw away energy.

Dropped at the turning point. Nothing is moving, so no energy is lost and no momentum argument is needed. The stored energy is still $\tfrac12(240)(0.100)^2 = 1.20$ J, and the amplitude is still 10.0 cm. But that energy now has to accelerate 0.90 kg: $\tfrac12(0.90)v_{\max}^2 = 1.20$, so $v_{\max}^2 = 2.40/0.90 = 2.667$ and $v_{\max} = 1.63$ m/s. Compare with 2.00 m/s before: heavier means slower for the same stored energy.

When energy leaks away

Real oscillators run down. Two kinds of loss show up constantly, and they behave completely differently.

Sliding friction takes a fixed amount per metre travelled. A kinetic friction force $f = \mu m g$ opposing the motion removes $f\,d$ of energy over a path length $d$, no matter how fast the block is going. That single sentence solves a whole family of problems.

Two consequences worth deriving. First, the block stops for good once its total energy has drained, so the total path length is $d_{\text{tot}} = E_0/(\mu m g)$ up to the small energy still stored where it finally stalls. Second, over one half-cycle the block travels from amplitude $A_n$ to amplitude $A_{n+1}$, a distance $A_n + A_{n+1}$, so

$\tfrac12 kA_n^2 - \tfrac12 kA_{n+1}^2 = \mu m g\,(A_n + A_{n+1})$

The difference of squares factors, the common $(A_n + A_{n+1})$ cancels, and what is left is startlingly simple:

$A_n - A_{n+1} = \frac{2\mu m g}{k}$

The amplitude falls by the same amount every half-swing. The envelope is a straight line, not the exponential you get from air drag. And the motion stops for good once the spring force can no longer beat static friction, which happens inside the band $|x| < \mu m g/k$.

Example 4. A 0.80 kg block on a spring with $k = 160$ N/m sits on a surface with $\mu = 0.20$, with $g = 9.8$ m/s$^2$. It is pulled 15.0 cm out and released. How many complete cycles does it manage?

The friction force is $f = \mu m g = (0.20)(0.80)(9.8) = 1.568$ N. Two lengths come from it. The half-width of the dead band is $x_0 = f/k = 1.568/160 = 0.0098$ m, and the amplitude loss per half-cycle is $2x_0 = 0.0196$ m.

So after $n$ half-swings the amplitude is $A_n = 0.150 - 0.0196n$. The block completes a half-swing only if it can reach the far side, which needs $A_{n+1} > 0$, that is $A_n > 0.0196$ m. Count:

$A_0 = 0.1500$, $A_1 = 0.1304$, $A_2 = 0.1108$, $A_3 = 0.0912$, $A_4 = 0.0716$, $A_5 = 0.0520$, $A_6 = 0.0324$, $A_7 = 0.0128$ m.

$A_6 = 0.0324$ m is bigger than 0.0196 m, so the seventh half-swing happens and lands at $A_7 = 0.0128$ m. But $A_7$ is smaller than 0.0196 m, so there is no longer enough stored energy to carry the block all the way across, and it stalls part-way through the eighth half-swing. Seven complete half-swings is three and a half swings, which is 3 complete cycles.

Air drag behaves differently: the loss per cycle is a fixed fraction of the energy, so $E$ decays exponentially. That is what a quality factor $Q$ encodes, through $E(t) = E_0 e^{-\omega_0 t/Q}$. Over one period $T = 2\pi/\omega_0$ the exponent becomes $-\omega_0 T/Q = -2\pi/Q$, so the fraction lost per cycle is

$\frac{\Delta E}{E} = 1 - e^{-2\pi/Q}$

For $Q = 40$: $2\pi/40 = 0.1571$, $e^{-0.1571} = 0.8546$, so the fork loses $1 - 0.8546 = 0.145$ of its energy each period. High $Q$ means a stingy oscillator.

When the moving thing is not one block

$K = \tfrac12 mv^2$ is a special case. Two upgrades cover most of what the problems throw at you, and both leave every other formula here intact: you only swap in a different effective mass.

If the oscillating object rolls without slipping, it stores kinetic energy in spin as well as in translation: $K = \tfrac12 mv^2 + \tfrac12 I\omega_{\text{spin}}^2$ with $\omega_{\text{spin}} = v/R$, which becomes $K = \tfrac12 m v^2 (1 + I/mR^2)$. A uniform solid cylinder has $I = \tfrac12 mR^2$, so the bracket is $1.5$ and $K = \tfrac34 mv^2$. For a 0.60 kg cylinder on a 180 N/m spring released at 9.0 cm: $\tfrac12(180)(0.090)^2 = 0.729$ J, and setting $\tfrac34(0.60)v^2 = 0.729$ gives $v^2 = 0.729/0.45 = 1.62$ and $v = 1.27$ m/s. Slower than a sliding block would be, because a third of the budget went into spin.

If two masses share one spring with the centre of mass at rest, the kinetic energy written in terms of the relative speed $v_{\text{rel}}$ is $K = \tfrac12 m_* v_{\text{rel}}^2$, where $m_* = m_1m_2/(m_1+m_2)$ is the reduced mass (a different $\mu$ is already doing duty as the friction coefficient, so it gets a star here). With 0.40 kg and 0.60 kg, $m_* = 0.24/1.00 = 0.24$ kg, and a 250 N/m spring stretched 5.0 cm gives $\tfrac12(0.24)v_{\text{rel}}^2 = \tfrac12(250)(0.050)^2$, so $v_{\text{rel}} = 0.050\sqrt{250/0.24} = 0.050 \times 32.27 = 1.61$ m/s.

One last idea, for the problem where the mass drifts slowly, like sand leaking from a swinging bucket. If a parameter changes over many periods, the ratio $E/\omega$ stays fixed. Since $E = \tfrac12 kA^2$ and $\omega = \sqrt{k/m}$, that forces $A^2 \propto \omega \propto m^{-1/2}$, so $A \propto m^{-1/4}$. Lose three quarters of the sand and the amplitude grows by $4^{1/4} = \sqrt2$.

Recognising that this is the tool

Reach for energy, not for $x(t)$, when a problem shows any of these:

• It asks for a speed at a stated position, or a position where the speed takes some value, and never mentions the time. That is $v = \omega\sqrt{A^2-x^2}$.
• It asks for an amplitude after something changed: putty added, a rider removed, drag applied. Find the energy just after the change and set it equal to $\tfrac12 kA'^2$.
• It gives a friction coefficient and asks for cycles, distance, or a final resting place. Energy in minus $\mu m g d$ out.
• It quotes a fraction of energy, a ratio of energies, or a $Q$. Ratios never need numbers for $m$ or $k$.
• It asks for a force at some point in the swing. Energy first for the speed, then $\sum F = mv^2/r$.

Time-domain methods answer "when". Energy answers everything else, and it does not care whether the restoring force came from a spring, from gravity on a pendulum, or from something stranger. That last point is the opening for the next lesson, where you stop assuming the potential is a parabola and start finding out what oscillates near the bottom of any well at all.