In the school laboratory a glass U-tube holds some water, with both surfaces level. Blow sharply into one arm and take your mouth away. The water surface in that arm drops, the other rises, and then both rock up and down, about once a second, for a while. Nothing in the tube looks like a spring. Yet the water keeps a steady rhythm, as a block on a spring does.
The weight of an unbalanced column
A U-tube is a bent tube of uniform bore, open at both ends, with its two arms upright. Let the liquid in it have density \(\rho\), and let \(A\) be the area of the bore. Measured along the tube, the liquid column has a total length \(L\). At rest the two surfaces are level, at the same height. That level is the mean position.
Now push the liquid so that the right surface rises by \(x\). The same volume leaves the left arm, so the left surface falls by \(x\). Compare the two sides at the height of the lower surface. Below that height, the liquid in the two arms balances. Above it, a column of height \(2x\) stands in the right arm with nothing to balance it. Its weight, \(\rho A(2x)g\), pushes the whole column back towards the level.
Apply the two steps. Step 1: the restoring force is \(F = -2\rho Agx\), so \(k_{\text{eff}} = 2\rho Ag\). Step 2: this force moves the whole liquid, which slides round the bend as one body. So \(m_{\text{eff}} = \rho AL\), the mass of all the liquid, not the mass of the column \(2x\). Then \(\omega^2 = 2\rho Ag/\rho AL = 2g/L\).
The density and the bore both cancelled. A denser liquid has a heavier unbalanced column, but it also has more mass to move, in the same proportion. A wider bore does the same. For the tube of Fig 6.1, \(L = 0.5\,\text{m}\), so \(T = 2\pi\sqrt{0.5/20} = 0.99\,\text{s}\). The restoring force at the largest displacement, \(5\,\text{cm}\), is \(4 \times 0.05 = 0.2\,\text{N}\).
A U-tube holds a column of water \(40\,\text{cm}\) long. The water is replaced by a column of mercury of the same length, \(13.6\) times as dense. What happens to the period?
The liquid column in a U-tube is \(40\,\text{cm}\) long, measured along the tube. Take \(g = 10\,\text{m/s}^2\). What is the period of its oscillation, in seconds?
A floating cylinder
A wooden cylinder floats upright in water. It sinks until the buoyant force, the weight of the water it pushes aside, equals its own weight. Let \(A\) be its cross-section and \(d\) the depth under water. Then \(\rho_w A d g = mg\), where \(\rho_w\) is the density of water. That floating position is the mean position.
Push the cylinder down by a further \(y\). A slice of height \(y\) goes under water, and the buoyant force grows by \(\rho_w A y g\). The weight has not changed, so this extra buoyancy is the net force, and it points up, back to the floating position. If the cylinder rises by \(y\), a slice comes out of the water and the buoyancy falls by the same amount. Now the net force points down. In both cases \(F = -\rho_w A g\,y\).
Step 1 gives \(k_{\text{eff}} = \rho_w A g\). Step 2: the force moves the cylinder, of mass \(m = \rho_w A d\) by the floating condition. So \(\omega^2 = \rho_w A g/\rho_w A d = g/d\), and \(T = 2\pi\sqrt{d/g}\). The period depends only on the depth at which the cylinder floats. For the cylinder of Fig 6.2, \(d = 0.12\,\text{m}\) and \(T = 2\pi\sqrt{0.012} = 0.69\,\text{s}\).
The period \(2\pi\sqrt{d/g}\) has the same form as \(2\pi\sqrt{e/g}\) for a hanging spring, and for the same reason. In both, the depth or the stretch at rest measures \(m/k\). It ignores the water that moves with the cylinder, so a real cylinder bobs a little more slowly.
A test tube, weighted with sand, floats upright in water with \(10\,\text{cm}\) under the surface. Take \(g = 10\,\text{m/s}^2\). (a) Find its period of bobbing. (b) It is moved to salt water of density \(1100\,\text{kg/m}^3\). Find the new period.
(a) \(T = 2\pi\sqrt{d/g} = 2\pi\sqrt{0.10/10} = 0.628\,\text{s}\).
(b) The denser liquid gives the same buoyancy with less depth: \(d = 10/1.1 = 9.09\,\text{cm}\). \(T = 2\pi\sqrt{0.0909/10} = 0.599\,\text{s}\).
The tube floats higher in salt water and bobs a little faster.
A log floats upright in a pond with \(8\,\text{cm}\) of it under the water. Take \(g = 10\,\text{m/s}^2\). What is the period of its small vertical oscillations, in seconds?
A ball in a tunnel through the Earth
Imagine a straight tunnel drilled through the centre of the Earth, from one side to the other, with the air pumped out. Drop a ball into one end. It falls towards the centre, faster and faster, passes it, and slows down as it rises towards the far end. There it stops, falls back, and returns. The motion is an oscillation. Is it SHM?
The chapter Gravitational Force and Field shows what gravity is inside a uniform Earth. At a distance \(r\) from the centre it is \(g\,r/R\), where \(R\) is the Earth's radius. Only the part of the Earth closer to the centre than the ball pulls on it. So gravity inside falls in proportion to the distance, from \(g\) at the surface to zero at the centre. Along the tunnel, with \(x\) measured from the centre, the force on the ball is \(F = -(mg/R)\,x\).
That is a linear restoring force, with \(k_{\text{eff}} = mg/R\) and \(m_{\text{eff}} = m\). So the ball moves in SHM with \(\omega^2 = g/R\), and \(T = 2\pi\sqrt{R/g}\). With \(R = 6.4 \times 10^6\,\text{m}\) and \(g = 9.8\,\text{m/s}^2\), \(T = 5080\,\text{s}\), which is \(84.6\) minutes. The ball reaches the far side in half of that, \(42\) minutes. At the centre it moves at \(\omega R = \sqrt{gR} = 7.9\,\text{km/s}\).
The tunnel need not pass through the centre. Drill it straight between two cities, along a chord. Measure \(x\) along the tunnel from its midpoint. Gravity points to the centre of the Earth, and its part along the tunnel is \(-(mg/R)\,x\) once again. So every straight tunnel, long or short, has the same period, \(84.6\) minutes. A trip from Delhi to Mumbai, about \(1150\,\text{km}\) apart, would take \(42\) minutes, with a top speed of about \(710\,\text{m/s}\) at the middle.
One straight tunnel runs through the centre of the Earth. A second, shorter straight tunnel joins two cities \(1000\,\text{km}\) apart. A ball is dropped into each, with no air or friction. How do the times to reach the far end compare?
A satellite skimming the Earth's surface, if there were no air, would circle once every \(2\pi\sqrt{R/g}\) as well, \(84.6\) minutes. The ball in the tunnel is the shadow of that satellite on a diameter. This is SHM as the projection of uniform circular motion, as in the chapter Kinematics of SHM.
Six springs, one line
Look back over the oscillators of this chapter so far. Each had its own cause of the restoring force. Yet each gave the same straight line through the origin, and the period came from the same two steps.
- A spring: \(k_{\text{eff}} = k\), \(m_{\text{eff}} = m\), \(T = 2\pi\sqrt{m/k}\).
- Gravity along an arc: \(k_{\text{eff}} = mg/l\), \(m_{\text{eff}} = m\), \(T = 2\pi\sqrt{l/g}\).
- A restoring torque: \(k_\theta = mgd\) or \(C\), inertia \(I\), \(T = 2\pi\sqrt{I/k_\theta}\).
- A liquid column: \(k_{\text{eff}} = 2\rho Ag\), \(m_{\text{eff}} = \rho AL\), \(T = 2\pi\sqrt{L/2g}\).
- Buoyancy: \(k_{\text{eff}} = \rho_w Ag\), \(m_{\text{eff}} = \rho_w Ad\), \(T = 2\pi\sqrt{d/g}\).
- Gravity inside the Earth: \(k_{\text{eff}} = mg/R\), \(m_{\text{eff}} = m\), \(T = 2\pi\sqrt{R/g}\).
The water in the laboratory U-tube now has its explanation. Blowing into one arm pushes the column aside and leaves an unbalanced column of height \(2x\). Its weight is the spring. A column of about \(50\,\text{cm}\) gives \(T = 2\pi\sqrt{0.5/20}\), about one second, the rhythm you saw. The water slows down and stops because of friction with the glass, which this lesson ignored.
People imagined a stone dropped into a well through the whole Earth long before anyone could calculate its motion. In 1966 the American physicist Paul Cooper wrote a short paper with the title "Through the Earth in Forty Minutes". He showed that a frictionless train in any straight tunnel between two cities would make the trip in about 42 minutes, powered by gravity alone. No such tunnel has been built. The heat and pressure deep inside the Earth make it impossible with any known material.
Early in the twentieth century the German engineer Hermann Frahm fitted ships with anti-roll tanks. Each is a large U-tube of water across the ship, from one side to the other. Its length is chosen so that the water oscillates with the same period as the ship's roll. As the ship rolls, the water sloshes from side to side a quarter of a cycle behind, and its weight works against the roll. Many cargo ships and ferries still carry such tanks.
- Using the mass of the unbalanced column as \(m_{\text{eff}}\) in a U-tube.The restoring force moves all the liquid, so \(m_{\text{eff}} = \rho AL\) with the whole length \(L\).
- Using the height of one arm as \(L\).\(L\) is the total length of the liquid column, measured along the tube through the bend.
- Using \(g\) for gravity inside the Earth.Inside a uniform Earth gravity falls to \(g\,r/R\). That is what makes the force proportional to the distance.