Press a steel ruler flat on a desk with most of it hanging over the edge. Bend the free end down and let go. The end flies up, stops, returns downward, and moves to and fro with a soft hum. Slide the ruler further onto the desk so that less of it hangs free, and twang it again. The hum rises: the short ruler vibrates faster. Tape a coin to the free end, and the vibration slows down.
What brings a body back
In the chapter Kinematics of SHM you described simple harmonic motion by its acceleration, \(a = -\omega^2 x\). Here \(x\) is the displacement from the mean position and \(\omega\) is the angular frequency. That chapter never asked what produces such an acceleration. Newton's second law says that an acceleration needs a force. So every body in SHM must feel a force, and this chapter finds it.
Think of the ruler at rest, straight. That position is the mean position, an equilibrium where the net force is zero. Bend the end down, and the bent steel pushes it back up. Bend it up, and the steel pushes it back down. On either side the force points towards the mean position. A force that always points back to the equilibrium is a restoring force.
A restoring force alone gives an oscillation, but not yet SHM. The body is pushed back, overshoots the mean position because it is moving, and is pushed back again from the other side. For the motion to be simple harmonic, the force must also grow in step with the displacement. Twice as far from the mean, twice the force.
A body has a linear restoring force when the net force on it is proportional to its displacement \(x\) from the mean position. The force points towards that position: \(F = -kx\). The positive constant \(k\) is the force constant.
Reading \(F = -kx\)
The constant \(k\) is the force constant of the oscillator, in N/m. It is the restoring force per metre of displacement, so it measures stiffness. For a spring it is the spring constant of the chapter on Work and Energy. For a pendulum, a floating log or a column of water, it will be something else, but it plays the same part. A large \(k\) means a strong pull back for a small displacement.
The minus sign carries the direction. Take \(x\) positive to the right of the mean position. At \(x = +3\,\text{cm}\) the force \(-kx\) is negative, so it points left, back to the mean. At \(x = -3\,\text{cm}\) the force is positive and points right, again back to the mean. At \(x = 0\) the force is zero. One formula with one sign covers both sides.
In the law \(F = -kx\) for a restoring force, what does the minus sign mean?
A body oscillates about a mean position. What must be true of the net force on it for the motion to be simple harmonic?
From the force to the period
Now combine the restoring force with Newton's second law. A body of mass \(m\) moves along the \(x\) axis, and the net force on it is \(-kx\). The acceleration is the force divided by the mass. Compare the result with the law of SHM from the chapter Kinematics of SHM, and the angular frequency appears.
\[ ma = -kx \]
Why. Newton's second law, with the net force equal to the restoring force \(-kx\).
\[ a = -\dfrac{k}{m}\,x \]
Why. divide both sides by the mass \(m\).
\[ \omega^2 = \dfrac{k}{m}, \qquad \omega = \sqrt{\dfrac{k}{m}} \]
Why. this has the form \(a = -\omega^2 x\), the law of SHM, so the motion is SHM with \(\omega^2\) equal to \(k/m\).
\[ T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{m}{k}} \]
Read it. the period and the angular frequency are related by \(\omega = 2\pi/T\).
Read the formula as a contest between two properties. The force constant \(k\) sits in the denominator of \(T\). A stiffer spring pulls harder at every displacement, so it gives larger accelerations and returns the body sooner. The mass sits in the numerator: a heavier body gets a smaller acceleration from the same force and takes longer. The ruler and the coin of the opening show both. The shorter ruler is stiffer, and the coin adds mass.
With \(1\,\text{kg}\) on the spring the period is \(2\pi\sqrt{1/10} = 1.99\,\text{s}\), and the graph fits four swings into \(8\,\text{s}\). With \(4\,\text{kg}\), four times the mass, it fits only two: the period is \(3.97\,\text{s}\), twice as long. With \(0.25\,\text{kg}\) the period halves to \(0.99\,\text{s}\). Watch the force arrow while you change the mass. At the same displacement the arrow has the same length for every block, because the spring does not know what it holds. Only the response to that force changes.
The mass of a block in a spring-block system is made four times as large. The spring is not changed. What happens to the period of oscillation?
A \(250\,\text{g}\) block oscillates on a smooth floor at the end of a spring of constant \(100\,\text{N/m}\). What is the angular frequency of the oscillation, in rad/s?
A large swing and a small swing
The formula \(T = 2\pi\sqrt{m/k}\) contains no amplitude. That is a strong claim. It says that a block pulled out \(25\,\text{cm}\) returns to the mean in exactly the same time as a block pulled out \(10\,\text{cm}\). Most students expect the larger swing to take longer, since the block has farther to go. Decide before you play the figure.
Two identical blocks sit on identical springs. The upper block is pulled \(10\,\text{cm}\) from its mean position and the lower one \(25\,\text{cm}\). Both are released together. Which block reaches the mean position first?
Play the figure and follow the dashed mean line. The two blocks cross it together, every time. The readouts show why. As the blocks pass the mean, the lower one moves at \(0.79\,\text{m/s}\) and the upper one at \(0.32\,\text{m/s}\), in the ratio \(2.5\) of their amplitudes. Everything about the lower motion is the upper motion scaled up by the same factor, positions, speeds and forces alike. A scaled copy of a motion keeps its timing.
For a linear restoring force \(F = -kx\), the period \(T = 2\pi\sqrt{m/k}\) is the same for every amplitude. An oscillator with this property is called isochronous.
This property has a name. An oscillator whose period does not depend on its amplitude is isochronous, "equal in time". It is why a spring-driven watch keeps time as its swing slowly fades. It is also a test of SHM. A ball rolling to and fro in a V-shaped groove feels a push of fixed size towards the bottom, not a push proportional to \(x\). Its period grows with the amplitude, so its motion is an oscillation but not SHM.
The amplitude does not change the period, but it changes almost everything else. Doubling the amplitude doubles the greatest speed, \(\omega A\), and the greatest force, \(kA\). Keep "how long" separate from "how fast".
A block on a spring is pulled out twice as far as before and released. Which statement is true?
One method for every oscillator
The rest of this chapter meets many oscillators: a block on a spring, a pendulum, a swinging metre stick and a twisting disc. There are also water in a U-tube, a floating cylinder and a ball in a tunnel through the Earth. Their forces come from springs, gravity, buoyancy and the weight of a liquid. Yet every one of them is solved by the same two steps.
Step 1: displace the body by a small \(x\) from its equilibrium and find the net restoring force; write it as \(F = -k_{\text{eff}}\,x\). Step 2: identify the inertia \(m_{\text{eff}}\) that this force accelerates; then \(\omega^2 = k_{\text{eff}}/m_{\text{eff}}\) and \(T = 2\pi\sqrt{m_{\text{eff}}/k_{\text{eff}}}\).
The constant \(k_{\text{eff}}\) is the effective force constant: whatever multiplies \(-x\) in the restoring force. It is often not a spring constant at all. The quantity \(m_{\text{eff}}\) is the inertia that the force has to move, often but not always the mass of the body. In the pendulums that twist, the "force" becomes a torque and the "mass" becomes a moment of inertia, yet the same two steps work.
A \(0.5\,\text{kg}\) block rests on a smooth floor, fixed to a wall by a spring. A spring balance shows that a pull of \(8\,\text{N}\) holds the block \(4\,\text{cm}\) from its mean position. Find the angular frequency, the period and the frequency of its oscillation.
Take \(x\) along the floor from the mean position. The spring force on the block is the only horizontal force.
Step 1: \(k_{\text{eff}} = F/x = 8/0.04 = 200\,\text{N/m}\), so \(F = -200x\). Step 2: the force moves the block alone, so \(m_{\text{eff}} = 0.5\,\text{kg}\).
\(\omega = \sqrt{200/0.5} = \sqrt{400} = 20\,\text{rad/s}\). Then \(T = 2\pi/20 = 0.314\,\text{s}\) and \(f = 1/T = 3.18\,\text{Hz}\).
The block makes about three swings every second. The amplitude was never needed.
An orbiting astronaut is weightless, so a weighing scale reads zero. Instead she sits on a chair of mass \(10\,\text{kg}\) mounted on springs of total constant \(2000\,\text{N/m}\). (a) What is the period of the empty chair? (b) With her seated, the period is \(1.20\,\text{s}\). What is her mass?
(a) \(T_0 = 2\pi\sqrt{10/2000} = 0.444\,\text{s}\).
(b) \(T = 2\pi\sqrt{m_{\text{total}}/k}\) gives \(m_{\text{total}} = kT^2/4\pi^2 = 2000 \times 1.44/39.48 = 73.0\,\text{kg}\). Her mass is \(73.0 - 10 = 63\,\text{kg}\).
Weight needs gravity, but inertia does not. The period measures inertia, so the method works where a scale cannot.
The ruler of the opening now makes sense. Its bent steel is a restoring force that grows with the bend, so the tip moves in SHM. Sliding the ruler onto the desk shortens the free part and makes it stiffer: \(k\) rises, and \(T\) falls, so the hum rises. The taped coin adds mass to the tip, so \(T\) grows and the hum falls. The size of the twang changes the loudness, not the note, because the period does not depend on the amplitude.
In 1676 the English scientist Robert Hooke printed a string of letters at the end of a book: "ceiiinosssttuv". It was an anagram that hid a discovery while claiming it first. Two years later he gave the solution in Latin, "ut tensio, sic vis": as the extension, so the force. That is the force law \(F = -kx\) of this lesson. Hooke used it to design springs for watches, the first isochronous oscillators small enough to carry.
- Thinking that a larger amplitude takes a longer time.For \(F = -kx\) the period is \(2\pi\sqrt{m/k}\), the same for every amplitude. A larger swing is faster, not slower.
- Writing \(T = 2\pi\sqrt{k/m}\).A heavier body is slower, so \(m\) goes on top: \(T = 2\pi\sqrt{m/k}\). It is \(\omega\) that equals \(\sqrt{k/m}\).
- Calling every oscillation simple harmonic.A restoring force alone gives an oscillation. SHM also needs the force to be proportional to the displacement.