Look at the back wheel of a motorcycle. Two springs, one on each side, hold the frame above the wheel. A bus has several leaf springs on each axle, and a mattress has hundreds of small coils. In each case many springs carry one load. When the load bounces, it does not bounce on any single spring. It bounces on all of them together.
Two springs, one displacement
In the chapter Potential Energy and Conservation of Energy you met springs in parallel. They are two springs side by side, both joined to the wall and to the block. Pull the block by \(x\), and each spring stretches by the same \(x\). Their forces add: \(-k_1x - k_2x = -(k_1 + k_2)x\). So step 1 gives \(k_{\text{eff}} = k_1 + k_2\), and the block oscillates as on one stiffer spring.
The test for parallel is not how the springs look but what they share. A block between two springs, one fixed to a wall on each side, looks like a chain. Move the block right by \(x\). The left spring stretches by \(x\) and pulls it left. The right spring is compressed by \(x\) and pushes it left. Both forces point back to the mean, and both are proportional to the same \(x\). So this pair is in parallel as well: \(k_{\text{eff}} = k_1 + k_2\).
A \(1\,\text{kg}\) block on a smooth floor lies between two walls. It is joined to the left wall by a spring of \(30\,\text{N/m}\) and to the right wall by a spring of \(70\,\text{N/m}\). What is the period of its oscillation, in seconds?
Two springs, one force
Now join the springs end to end: wall, spring 1, spring 2, block. These are springs in series. The junction between them is light, so the net force on it is zero. The two springs therefore pull with the same force \(F\). Spring 1 stretches by \(F/k_1\) and spring 2 by \(F/k_2\). The block moves by the sum of the two stretches.
\[ x = x_1 + x_2 \]
Why. the block moves by the stretch of spring 1 plus the stretch of spring 2.
\[ x = \dfrac{F}{k_1} + \dfrac{F}{k_2} \]
Why. the light junction carries no net force, so the same force \(F\) acts in both springs; each stretches by \(F/k\).
\[ \dfrac{x}{F} = \dfrac{1}{k_1} + \dfrac{1}{k_2} \]
Why. divide by \(F\).
\[ \dfrac{1}{k_s} = \dfrac{1}{k_1} + \dfrac{1}{k_2} \]
Read it. one spring that stretches by \(x\) under \(F\) has constant \(k_s = F/x\).
The series pair is softer than either spring alone. For two equal springs, \(k_s = k/2\): the same force now stretches twice the length of spring. A softer spring gives a longer period. The parallel pair of two equal springs has \(k_p = 2k\). Between the two arrangements of the same two springs, the period changes by a factor of \(\sqrt{4} = 2\).
In Fig 3.1 the three lines of the graph fan out from the origin. The series line is the gentlest, because a given displacement brings the smallest force. The parallel line is the steepest. A steeper line pulls harder at every \(x\), so the block returns sooner. In the series chip, watch the black junction dot. It moves exactly half as far as the block, because two equal springs share the stretch equally.
Springs of \(100\,\text{N/m}\) and \(300\,\text{N/m}\) hold a \(3\,\text{kg}\) block on a smooth floor. Find the period (a) when the springs are in series, and (b) when they are in parallel.
(a) \(1/k_s = 1/100 + 1/300 = 4/300\), so \(k_s = 75\,\text{N/m}\). \(T = 2\pi\sqrt{3/75} = 2\pi \times 0.2 = 1.26\,\text{s}\).
(b) \(k_p = 100 + 300 = 400\,\text{N/m}\). \(T = 2\pi\sqrt{3/400} = 0.544\,\text{s}\).
(c) The series pair is softer than even the \(100\,\text{N/m}\) spring, and the parallel pair is stiffer than the \(300\,\text{N/m}\) spring.
A block oscillates on one spring. A second, identical spring is joined to the end of the first, in series, and the block is fixed to the far end. What happens to the period?
Cutting a spring
A uniform spring of length \(L\) can be seen as two shorter springs in series. Under a force \(F\), every centimetre of it stretches by the same amount. So a piece of length \(L_1\) stretches by the fraction \(L_1/L\) of the whole stretch. Its constant is therefore larger by the factor \(L/L_1\). This is the rule \(k_1L_1 = kL\) from the chapter Potential Energy and Conservation of Energy.
Apply it to the period. A block on half the spring has twice the force constant, so its period is \(1/\sqrt{2}\) as long. Cut a spring into four equal pieces and fit all four side by side, in parallel. Each piece has \(4k\), and four of them in parallel give \(16k\). The period falls to a quarter of the period on the whole spring.
A block on a uniform spring oscillates with a period of \(2\,\text{s}\). The spring is cut into four equal pieces, and the block is hung on all four together, side by side. What is the new period, in seconds?
Two blocks on one spring
Now let neither end of the spring be fixed. Two blocks, of masses \(m_1\) and \(m_2\), lie on a smooth floor joined by a spring of constant \(k\). Stretch the spring and release both blocks. No outside force acts along the floor, so the total momentum stays zero, and the centre of mass stays still. The blocks move in opposite directions, the lighter one farther.
Look at the point C in Fig 3.2. It sits on the spring above the centre of mass, and it never moves. So each block is really a block on its own fixed spring: the piece between it and C. The piece on the side of \(m_1\) has length \(L\,m_2/(m_1 + m_2)\), here \(1.0\,\text{m}\). By the cut-spring rule its constant is \(k(m_1 + m_2)/m_2 = 9\,\text{N/m}\). Then \(T = 2\pi\sqrt{m_1/k_1} = 2\pi\sqrt{1/9} = 2.09\,\text{s}\). The other block gives the same period, as it must.
\[ L_1 = \dfrac{m_2}{m_1 + m_2}\,L \]
Why. the still point C is the centre of mass; its distances from the blocks are in the inverse ratio of their masses.
\[ k_1 = \dfrac{kL}{L_1} = \dfrac{k(m_1 + m_2)}{m_2} \]
Why. the piece between \(m_1\) and C is a cut spring, with \(k_1L_1 = kL\).
\[ T = 2\pi\sqrt{\dfrac{m_1}{k_1}} = 2\pi\sqrt{\dfrac{m_1m_2}{(m_1 + m_2)\,k}} \]
Why. block 1 is a block on a fixed spring of constant \(k_1\).
\[ \mu = \dfrac{m_1m_2}{m_1 + m_2}, \qquad T = 2\pi\sqrt{\dfrac{\mu}{k}} \]
Read it. name the combination of masses the reduced mass \(\mu\).
The reduced mass is smaller than either mass. For the blocks of Fig 3.2 it is \(2/3\,\text{kg}\). It behaves sensibly at the limits. If \(m_2\) becomes very large, \(\mu\) becomes \(m_1\). A very heavy block barely moves, like a wall, and the light block oscillates on the whole spring. If the two masses are equal, \(\mu = m/2\), and C sits at the middle of the spring.
Trolleys of \(2\,\text{kg}\) and \(3\,\text{kg}\) on a smooth track are joined by a spring of constant \(120\,\text{N/m}\). The spring is compressed and the trolleys are released. Find the period of their oscillation.
No outside force acts along the track, so the centre of mass stays still and both trolleys oscillate with one period.
\(\mu = m_1m_2/(m_1 + m_2)\) and \(T = 2\pi\sqrt{\mu/k}\).
\(\mu = 2 \times 3/5 = 1.2\,\text{kg}\). \(\omega = \sqrt{120/1.2} = 10\,\text{rad/s}\), so \(T = 2\pi/10 = 0.628\,\text{s}\).
Both trolleys swing with the period \(0.63\,\text{s}\), the lighter one with the larger amplitude, in the ratio \(3 : 2\).
Two identical \(1\,\text{kg}\) blocks on a smooth floor are joined by a spring of constant \(50\,\text{N/m}\). The spring is stretched and the blocks are released. What is the angular frequency of their oscillation, in rad/s?
Return to the motorcycle of the opening. Its two shock absorbers both join the frame to the same axle, so they share one displacement: they are in parallel, and their constants add. A rider bounces on one spring of the combined constant, with the period \(2\pi\sqrt{m/(k_1 + k_2)}\). The leaf springs of a bus and the coils of a mattress are larger sets of the same kind. Each set, however many springs it has, can be replaced by one spring.
A hydrogen molecule is two atoms joined by a chemical bond that acts, for small stretches, like a spring. Its vibration frequency is set by the bond's stiffness and the reduced mass of the two atoms. In heavy hydrogen each atom has twice the mass, while the bond, made by the same electrons, has the same stiffness. The reduced mass doubles, so the molecule vibrates at \(1/\sqrt{2}\), about \(0.71\), of the frequency. Chemists use such shifts in infrared light to tell the two apart.
- Adding the constants of any two springs.Constants add only when the springs share one displacement (parallel). When they carry one force (series), the reciprocals add.
- Calling a block between two opposite springs "series".Both springs change length by the same displacement and both push the block back, so they are in parallel.
- Using one block's mass for two blocks on a free spring.Neither end of the spring is fixed. Use the reduced mass \(\mu = m_1m_2/(m_1 + m_2)\).