In 1790 the scientists of France looked for a natural length to call the metre. One proposal was the length of a seconds pendulum. Such a pendulum takes one second for each swing, so its period is \(T = 2\,\text{s}\). The chapter Dynamics of SHM gives the period as \(T = 2\pi\sqrt{l/g}\). That formula gives the period from the length. The scientists needed the opposite: the length from the period.
An equation is a balance
An equation says that two quantities are equal. If both sides are changed in the same way, they stay equal. You may add the same quantity to both sides, or subtract it. You may multiply or divide both sides by the same quantity, provided that quantity is not zero. You may square both sides, or take the square root of both, keeping the root that makes physical sense. Every rearrangement is a chain of such steps.
Undo in the reverse order
To make a quantity the subject, look at what was done to it, and in what order. Then undo those operations in the reverse order, the last one first. In \(T = 2\pi\sqrt{l/g}\), the length \(l\) was first divided by \(g\). Then the square root was taken. Last, the result was multiplied by \(2\pi\). So divide by \(2\pi\) first, then square, and multiply by \(g\) last. Fig 1.1 shows the two chains side by side.
To make a quantity the subject of a formula, apply the same operation to both sides, one step at a time. Undo the operations acting on that quantity in the reverse of the order in which they were applied. Never divide by a quantity that can be zero. After taking a square root, keep the sign that the physics allows.
\[ T = 2\pi\sqrt{\dfrac{l}{g}} \]
Why. the period of a simple pendulum swinging through small angles. The length \(l\) sits under three operations.
\[ \dfrac{T}{2\pi} = \sqrt{\dfrac{l}{g}} \]
Why. the last operation was the multiplication by \(2\pi\), so it is undone first. Divide both sides by \(2\pi\).
\[ \dfrac{T^2}{4\pi^2} = \dfrac{l}{g} \]
Why. squaring undoes the square root. The whole left side is squared, so \(2\pi\) becomes \(4\pi^2\).
\[ l = \dfrac{gT^2}{4\pi^2} \]
Read it. the first operation was the division by \(g\), so it is undone last. Multiply both sides by \(g\).
Now, and only now, substitute the numbers. With \(T = 2\,\text{s}\) and \(g = 9.8\,\text{m/s}^2\), \(l = 9.8 \times 4/(4\pi^2) = 9.8/\pi^2 = 0.993\,\text{m}\). The seconds pendulum is very nearly one metre long, which is why it was proposed. The units agree as well: \(\text{m/s}^2 \times \text{s}^2 = \text{m}\), a length.
A formula is also evaluated in a fixed order. Brackets come first, then powers and roots, then multiplication and division, and last addition and subtraction. Take \(s = ut + \tfrac12 at^2\) with \(u = 5\,\text{m/s}\), \(t = 4\,\text{s}\) and \(a = 10\,\text{m/s}^2\). The power comes first, \(4^2 = 16\). Then the two products, \(5 \times 4 = 20\) and \(\tfrac12 \times 10 \times 16 = 80\). The sum comes last, \(s = 100\,\text{m}\). A rearrangement walks this order backwards.
Make the acceleration \(a\) the subject of \(v = u + at\).
In a school laboratory a pendulum \(0.50\,\text{m}\) long makes \(10\) complete swings in \(14.2\,\text{s}\). Rearrange \(T = 2\pi\sqrt{l/g}\) to find \(g\), in m/s².
Symbols first, numbers last
Keep the letters until the last line, for three reasons. First, some quantities cancel, and a cancelled quantity needs no value: the mass of a pendulum bob never enters its period. Second, the arithmetic is done once, with one rounding at the end. Third, the unit of the answer can be checked. A rearranged formula whose unit comes out wrong has a mistake in it.
Units travel with the numbers
Write each number with its unit, and let the units multiply and divide like the numbers. A car moving at \(20\,\text{m/s}\) brakes at \(5\,\text{m/s}^2\) and stops. From \(v^2 = u^2 + 2as\), the distance is \(s = (v^2 - u^2)/2a\). With direction of motion positive, \(s = (0 - 400\,\text{m}^2/\text{s}^2)/(2 \times (-5\,\text{m/s}^2)) = 40\,\text{m}\). The units give \(\text{m}^2/\text{s}^2 \div \text{m/s}^2 = \text{m}\), a length, as a distance must be.
A student makes \(l\) the subject of \(T = 2\pi\sqrt{l/g}\) and writes \(l = 4\pi^2/(gT^2)\). What shows at once that this is wrong?
A bus moving at \(30\,\text{m/s}\) on a highway brakes with a constant deceleration of \(6\,\text{m/s}^2\). Rearrange \(v^2 = u^2 + 2as\) and find its stopping distance, in metres.
When the unknown appears twice
Sometimes the wanted quantity appears on both sides of the equation. Then no single chain of undoing will free it. Four steps do. Expand the brackets. Bring every term that contains the unknown to one side and every other term to the other side. Take the unknown out as a common factor. Divide by what multiplies it.
A mixture of hot and cold water is the standard case. A hot body of heat capacity \(C_1\) at \(T_1\) and a cold body of heat capacity \(C_2\) at \(T_2\) reach one final temperature \(T\). The heat lost by the hot body equals the heat gained by the cold one, and \(T\) appears on both sides. One step below is left for you.
\[ C_1(T_1 - T) = C_2(T - T_2) \]
Why. the heat lost by the hot body equals the heat gained by the cold body.
\[ C_1T_1 - C_1T = C_2T - C_2T_2 \]
Why. expand both brackets.
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Your turn. which line follows? Collect the terms in \(T\) on one side and take \(T\) out as a factor.
\[ T = \dfrac{C_1T_1 + C_2T_2}{C_1 + C_2} \]
Read it. divide both sides by \(C_1 + C_2\), which is never zero.
The result says that \(T\) is a weighted mean of the two temperatures. Each temperature is weighted by its heat capacity. In the figure below the two bodies sit on a beam at their own temperatures, and the beam balances at \(T\). Move the slider and watch the balance point move towards the larger heat capacity.
\(200\,\text{g}\) of water at \(90\,\text{°C}\) is mixed with \(100\,\text{g}\) of water at \(30\,\text{°C}\) in an insulated flask. Find the final temperature. Take \(c = 4200\,\text{J/(kg K)}\).
Hot water: \(C_1 = 0.2 \times 4200 = 840\,\text{J/K}\) at \(T_1 = 90\,\text{°C}\). Cold water: \(C_2 = 0.1 \times 4200 = 420\,\text{J/K}\) at \(T_2 = 30\,\text{°C}\).
Heat lost equals heat gained: \(840(90 - T) = 420(T - 30)\).
\(75600 - 840T = 420T - 12600\), so \(88200 = 1260\,T\) and \(T = 70\,\text{°C}\).
The mixture settles at \(70\,\text{°C}\). Check: the hot water loses \(840 \times 20 = 16800\,\text{J}\) and the cold gains \(420 \times 40 = 16800\,\text{J}\).
\(300\,\text{g}\) of water at \(80\,\text{°C}\) is poured into \(100\,\text{g}\) of water at \(20\,\text{°C}\) in an insulated jug. What is the final temperature, in °C?
Identities worth knowing
An identity is an equation that is true for every value of its letters. A few identities appear again and again in physics. The first three are used most often.
\[ (a \pm b)^2 = a^2 \pm 2ab + b^2, \qquad a^2 - b^2 = (a - b)(a + b), \qquad (x + a)(x + b) = x^2 + (a + b)x + ab \]
The cubes follow the same pattern. \((a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3\), and \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\). Rarely, \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\) also helps.
The difference of two squares is the one physics uses most. In an elastic collision on a line, energy gives \(m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)\), and momentum gives \(m_1(u_1 - v_1) = m_2(v_2 - u_2)\). Factor each difference of squares and divide the energy equation by the momentum equation. What is left is \(u_1 + v_1 = v_2 + u_2\), an equation with no squares at all. The chapter Collisions uses exactly this step.
Use an identity to find \(103 \times 97\) without long multiplication.
Three slips break most rearrangements. Dividing by a quantity that can be zero: \(t = (v - u)/a\) says nothing when \(a = 0\), because then \(v = u\) at every time. Squaring both sides: \(v = -3\) and \(v = 3\) give the same \(v^2 = 9\), so a square root must be followed by a choice of sign. Taking a root term by term: \(\sqrt{a^2 + b^2}\) is not \(a + b\), since \(\sqrt{9 + 16} = 5\), not \(7\).
Every formula of the book is rearranged at some point. The third equation of motion gives a braking distance. The period of a pendulum gives the value of \(g\) in the chapter Dynamics of SHM. The principle of calorimetry gives the final temperature of a mixture, with the unknown on both sides. Young's modulus of a wire and the energy stored in a stretched wire are found by rearranging. So is the pressure across the wall between two bubbles. The same step finds the point where the pulls of the Earth and the Moon cancel. It also finds where a third charge feels no force, and the wavelength of an electron of known energy.
- Dividing only one term by a quantity.In \(v - u = at\), divide the whole of \(v - u\) by \(t\): \(a = (v - u)/t\), not \(v/t - u\).
- Squaring only part of one side.Square the whole side. \((T/2\pi)^2 = T^2/4\pi^2\); the \(2\pi\) is squared too.
- Substituting numbers too early.Rearrange with letters first. Substitute the numbers, with their units, in the last line.