Spring-Mass Systems — padho.wiki

In short

A mass m attached to a spring of stiffness k executes simple harmonic motion with angular frequency \omega = \sqrt{k/m} and time period

T = 2\pi\sqrt{\frac{m}{k}}.

Hanging the mass vertically shifts the equilibrium position downward by x_0 = mg/k, but the time period is unchanged — gravity does not appear in the formula. When two springs are connected to the same mass in parallel (both resist the same displacement), the effective stiffness adds: k_\text{eff} = k_1 + k_2. When they are connected in series (one after the other, the same force passes through both), the compliances add: 1/k_\text{eff} = 1/k_1 + 1/k_2. A mass between two fixed walls, connected by a spring on each side, behaves like a parallel combination: k_\text{eff} = k_1 + k_2. Every spring-mass problem in mechanics reduces to these rules — identify the effective spring constant, plug into T = 2\pi\sqrt{m/k_\text{eff}}, and you are done.

Ride a Maruti Alto over a Mumbai pothole. For a couple of seconds after the jolt, the whole car bounces up and down — boing, boing, boing — and then settles. The body of the car has a mass of about a tonne. Under each wheel there is a coiled steel spring stiff enough to hold up that weight and still flex a few centimetres under a bump. What you feel in that bounce is a spring-mass oscillator: a thousand kilograms bobbing on four springs at a frequency you could clock with a stopwatch.

Or walk onto a trampoline at a Jaipur amusement park. Your own mass — call it 50 kg — is held up by a mesh of springs around the rim. Give a little hop and you bounce at a steady rhythm. Increase your mass (ask a friend to join you) and the rhythm slows down. Make the springs stiffer (a competition trampoline has much tighter springs than a child's one) and the rhythm speeds up. You are, without thinking about it, doing an experiment in spring-mass physics.

Every example of a block on a spring — a shock absorber on an autorickshaw, the coiled spring inside a ballpoint pen, the suspension of a Delhi Metro carriage, the diaphragm of a loudspeaker in your headphones — obeys the same equation and the same time period. In the previous two chapters you met simple harmonic motion and its solution x(t) = A\sin(\omega t + \varphi). This chapter takes the cleanest physical example of SHM — a mass attached to a spring — and works through it completely: the horizontal case, the vertical case, springs in series, springs in parallel, a mass sandwiched between two walls by two springs. By the end you will be able to look at almost any spring configuration, find the effective spring constant in your head, and write down the period.

The horizontal spring-mass — starting clean

Put a mass m on a frictionless horizontal surface. Attach it to a wall by a spring of stiffness k and natural length L_0. Let x(t) be the displacement of the mass from the position where the spring is at its natural length — the equilibrium. When x > 0 the spring is stretched and pulls the mass back toward equilibrium; when x < 0 it is compressed and pushes the mass back. Hooke's law gives the spring force exactly:

F_\text{spring} = -kx.

A block of mass $m$ on a frictionless surface, connected to a wall by a spring of stiffness $k$. Displacement $x$ is measured from the equilibrium position (where the spring is at its natural length). The spring force is $F = -kx$ — always pointing back to equilibrium.

Apply Newton's second law to the block. The only horizontal force is the spring force, so

m \frac{d^2 x}{dt^2} = -k x.

Rearranging,

\frac{d^2 x}{dt^2} = -\frac{k}{m}\, x = -\omega^2 x, \qquad \omega = \sqrt{\frac{k}{m}}. \tag{1}

Why: the ratio k/m has units of 1/\text{s}^2 (since k is in N/m = kg/s² and m is in kg), so \sqrt{k/m} has units of 1/\text{s} — the right units for an angular frequency. A stiffer spring increases \omega; a heavier mass decreases it.

This is the equation of SHM, already derived in Equation of SHM and Phase. Its general solution is

x(t) = A\sin(\omega t + \varphi),

with amplitude A and phase \varphi set by initial conditions. The period is

\boxed{\; T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}. \;} \tag{2}

That is the entire result for a horizontal spring-mass. Three things are worth staring at before moving on.

First: the period depends only on the two numbers m and k. Not the amplitude — a block displaced by 10 cm takes exactly the same time to complete one oscillation as a block displaced by 1 cm. Not any external force — gravity, friction (if there were any), the colour of the block — none of them appear. This is isochronism, the same property Galileo noticed in the swinging chandelier at Pisa. Small swings and big swings take the same time.

Second: doubling the mass multiplies the period by \sqrt{2} \approx 1.414. Doubling the stiffness divides it by \sqrt{2}. Quadrupling either one doubles or halves the period. This is the scaling law — and it is the thing you should be able to quote at a glance.

Third: the formula has the right dimensions. [m] = \text{kg}, [k] = \text{N/m} = \text{kg/s}^2, so m/k has units of s², and its square root is in seconds. Any time you are unsure of the formula, a quick dimensional check catches an error.

A 1 kg mass on a 25 N/m spring has $\omega = \sqrt{25/1} = 5$ rad/s and $T = 2\pi/5 \approx 1.257$ s. Released from $x = 0.15$ m with zero velocity, it traces out $x(t) = 0.15 \cos(5t)$. Ghost markers at the quarter, half, three-quarter, and full period fall at $0$, $-A$, $0$, $+A$. Click replay to watch again.

Explore how T depends on m and k

The algebra says T = 2\pi\sqrt{m/k}. Dragging mass and stiffness lets you feel how each one shifts the period.

Drag the red handle along the $x$-axis to change the mass, and the dark handle along the $y$-axis to change the spring constant. The curve $T = 2\pi\sqrt{m/k}$ updates live. Notice that quadrupling $m$ doubles $T$ (square root), while quadrupling $k$ halves $T$.

Vertical spring — gravity shifts the equilibrium, not the period

Hang the same mass m from the same spring k, now vertically. Gravity pulls the mass down by mg; the spring stretches until the spring force balances gravity. Let y be the position of the mass measured downward from the natural-length position.

At equilibrium, the net force is zero:

k y_0 = m g \quad\Rightarrow\quad y_0 = \frac{m g}{k}. \tag{3}

Why: the spring pulls up with force k y_0 (because it is stretched by y_0 below its natural length), and gravity pulls down with force m g. At equilibrium these cancel, giving y_0 = m g / k.

Now displace the mass from this equilibrium. Let \eta = y - y_0 be the displacement from the new equilibrium. The net force on the mass is spring force minus gravity:

F_\text{net} = -k y + m g = -k (y_0 + \eta) + m g = -k y_0 + m g - k \eta.

But k y_0 = m g from equation (3), so -k y_0 + m g = 0. Therefore:

F_\text{net} = -k \eta. \tag{4}

Why: gravity's constant pull shifts where the equilibrium is, but around that shifted equilibrium the restoring force is still -k times the displacement. The "gravity" and the "static stretch of the spring" cancel out exactly — leaving pure SHM in the variable \eta.

Newton's second law gives

m \frac{d^2 \eta}{dt^2} = -k \eta,

which is the same equation (1) as before. The period is therefore the same:

T = 2\pi\sqrt{\frac{m}{k}}. \tag{5}

Gravity does not appear in T.

Hanging the mass shifts the equilibrium down by $y_0 = mg/k$. Measured from the new equilibrium (the middle panel), the restoring force is again $F = -k\eta$ — gravity has disappeared. The oscillation about this shifted equilibrium has exactly the same period $T = 2\pi\sqrt{m/k}$ as the horizontal case.

This is a small pedagogical miracle. The complication you might have feared — "gravity must change something" — doesn't change the period at all. It shifts where the mass oscillates, but not how fast. The reason is exactly the linearity of Hooke's law. In a nonlinear spring, gravity would change the period (because the effective stiffness at the shifted equilibrium would differ from the stiffness at the natural length). For the ideal Hookean spring, the two effects decouple perfectly.

A practical consequence: the formula gives you a quick way to measure k. Hang the mass, measure the static stretch y_0, and read off k = mg/y_0. Then from k and m predict the period. Most laboratory spring-constant experiments use exactly this trick.

Combining springs — the rules every problem reduces to

Real systems rarely have a single spring. A car has four shock absorbers. A bed has dozens of parallel coils. A torsional suspension has two springs, one on each side of the driveshaft. Fortunately, any configuration of ideal springs attached to one mass reduces to a single effective spring with an effective stiffness k_\text{eff}. The period is then T = 2\pi\sqrt{m/k_\text{eff}}. The two elementary combinations are series and parallel.

Springs in parallel — the stiffnesses add

Two springs are in parallel if they share the same displacement — one end of each is fixed to the same wall and the other end to the same mass.

Two springs in parallel: both anchored to the same wall, both attached to the same block. A displacement $x$ stretches *both* springs by the same amount $x$, so each exerts its own restoring force, and the two forces add.

Displace the mass by x. Spring 1 stretches by x and exerts force -k_1 x. Spring 2 also stretches by x and exerts force -k_2 x. The total force on the mass is the sum:

F_\text{net} = -k_1 x - k_2 x = -(k_1 + k_2)\, x.

This is exactly Hooke's law with an effective spring constant

\boxed{\; k_\text{eff} = k_1 + k_2 \quad\text{(parallel)}. \;} \tag{6}

Why: parallel springs share displacement but add force. Each spring has the same x, and each contributes its own -kx to the total. Adding forces means adding spring constants.

The same rule extends to any number of springs in parallel:

k_\text{eff} = k_1 + k_2 + k_3 + \cdots.

Springs in series — the compliances add

Two springs are in series if they are joined end to end — one end of spring 1 to the wall, the other to spring 2, the other end of spring 2 to the mass. The same force passes through both springs.

Two springs in series: joined end-to-end. The same internal force $T$ pulls through both springs; each stretches by its own amount, and the total stretch is $x = x_1 + x_2$.

Pull the mass by a total displacement x. Let spring 1 stretch by x_1 and spring 2 stretch by x_2, so x = x_1 + x_2. Because spring 2's left end is not attached to anything rigid (it is attached to spring 1's right end), the force must be the same throughout: the tension T that spring 1 pulls with at its right end must equal the tension that spring 2 pulls with at its left end. Otherwise the massless join point would have a nonzero net force and infinite acceleration.

So:

T = k_1 x_1 = k_2 x_2.

Solve for the individual stretches:

x_1 = \frac{T}{k_1}, \qquad x_2 = \frac{T}{k_2}.

Add them:

x = x_1 + x_2 = T\left(\frac{1}{k_1} + \frac{1}{k_2}\right).

The effective spring constant is the one that satisfies T = k_\text{eff} \, x, so

\frac{1}{k_\text{eff}} = \frac{x}{T} = \frac{1}{k_1} + \frac{1}{k_2}.

\boxed{\; \frac{1}{k_\text{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \quad\text{(series)}. \;} \tag{7}

Why: series springs share force but add displacement. Because the total stretch is the sum of individual stretches, and each individual stretch is inversely proportional to that spring's stiffness, the inverses add. The effective spring is always softer than the softest individual spring — the weakest link dominates.

Equivalently, for two springs:

k_\text{eff} = \frac{k_1 k_2}{k_1 + k_2},

the product-over-sum rule. For k_1 = k_2 = k this gives k_\text{eff} = k/2: two identical springs in series are half as stiff as one. And T goes up by \sqrt{2}.

A memory trick: parallel is like wires, series is like resistors

The rule is easy to confuse with the electrical analogues. The memory trick:

Springs in parallel add like resistors in series. (Both add stiffness / resistance.)
Springs in series add like resistors in parallel. (Both add compliance / conductance.)

Mechanical springs and electrical resistors share the structure of the math but with roles swapped. Parallel springs are stiffer (the wall has two handles pulling back). Series springs are softer (the wall's pull has to travel through a long chain).

Mass between two walls — two springs, one on each side

A common JEE-style configuration: a mass sits between two fixed walls, connected to the left wall by spring k_1 and to the right wall by spring k_2. Is this parallel or series?

A block sandwiched between two walls by two springs. Displace the block to the right by $x$: spring $k_1$ stretches by $x$, spring $k_2$ compresses by $x$. Both springs pull the block leftward.

Displace the mass to the right by x. Spring k_1 (on the left) is now stretched by x — it pulls the mass leftward with force k_1 x. Spring k_2 (on the right) is now compressed by x — it pushes the mass leftward with force k_2 x. Both forces point in the same direction (opposite to the displacement), so they add:

F_\text{net} = -k_1 x - k_2 x = -(k_1 + k_2)\, x.

This is the parallel rule. So:

k_\text{eff} = k_1 + k_2, \qquad T = 2\pi\sqrt{\frac{m}{k_1 + k_2}}. \tag{8}

Why: although the springs are on opposite sides of the mass, they both oppose the displacement. One spring's stretching and the other's compressing both create a restoring force in the same direction — so the stiffnesses add, just as they do for two parallel springs on one side.

The reader who at first wants to call this "series" should stare at the picture. The defining question is not the geometry of where the springs are attached; it is whether the two springs share displacement (parallel) or share force (series). In the walls-and-mass setup, both springs stretch/compress by the same amount — the same displacement of the mass — so they are in parallel.

Worked examples

Example 1: The Maruti Alto over a Mumbai pothole

A Maruti Alto has a body mass of roughly 800 kg distributed over four wheels. Each of the four suspension springs has a stiffness k = 35{,}000 N/m (a typical value for a small car). Treat the car body as bouncing on the four springs in parallel, and estimate:

(a) the effective spring constant of the suspension, (b) the time period of the vertical bounce, (c) the frequency in hertz.

The Alto body on four identical springs, one per wheel. A vertical bounce displaces the body downward by $x$; *all four* springs compress by $x$, so they share displacement — the parallel rule applies.

Step 1. Identify the spring combination.

All four springs compress (or stretch) by the same amount when the car body bounces vertically — they share the displacement. So they are in parallel.

Why: each spring's top is attached to the same (rigid) car body, and each spring's bottom is attached to the ground. When the body moves down by x, each spring shortens by x — the same x for all four. Shared displacement means parallel.

Step 2. Apply the parallel rule.

k_\text{eff} = k_1 + k_2 + k_3 + k_4 = 4k = 4 \times 35{,}000 = 140{,}000 \text{ N/m}.

Why: for n identical parallel springs of stiffness k each, k_\text{eff} = n k. Four springs make the body feel four times as stiffly supported as a single spring would.

Step 3. Find the time period.

T = 2\pi\sqrt{\frac{M}{k_\text{eff}}} = 2\pi\sqrt{\frac{800}{140{,}000}} = 2\pi\sqrt{0.00571}\; \text{s}.

Compute: \sqrt{0.00571} = 0.0756. So

T = 2\pi \times 0.0756 \approx 0.475 \text{ s}.

Why: plug directly into T = 2\pi\sqrt{M/k_\text{eff}}. The mass is 800 kg (the full body; the springs all share its weight), the effective stiffness is 140,000 N/m.

Step 4. Find the frequency.

f = \frac{1}{T} = \frac{1}{0.475} \approx 2.1 \text{ Hz}.

Result: k_\text{eff} = 140{,}000 N/m, T \approx 0.47 s, f \approx 2.1 Hz.

What this shows: The Alto's body bounces at about 2 cycles per second. This is — not coincidentally — the typical bounce frequency of every passenger car, because the human body is most comfortable with vertical oscillations around 1–1.5 Hz, and designers tune the suspension to hit that range. The real suspension has dampers (shock absorbers) that kill the oscillation within a bounce or two; without them, the car would hop like a pogo stick every time you hit a pothole on the Mumbai–Pune expressway. Those dampers are what turn this undamped SHM into the damped oscillation you will meet in the next chapter.

Example 2: Two springs in series and in parallel

A 2 kg block can be attached to two springs of stiffnesses k_1 = 100 N/m and k_2 = 300 N/m. Find the time period of the block's oscillation in two configurations:

(a) both springs in parallel (both attached to the same wall, both to the block), (b) both springs in series (wall — spring 1 — spring 2 — block).

The same two springs in two arrangements. Parallel: both anchored to the wall, both to the block. Series: one after the other.

Step 1 (parallel). Add the stiffnesses.

k_\text{eff}^\text{par} = k_1 + k_2 = 100 + 300 = 400 \text{ N/m}.

Step 2 (parallel). Compute the period.

T_\text{par} = 2\pi\sqrt{\frac{m}{k_\text{eff}^\text{par}}} = 2\pi\sqrt{\frac{2}{400}} = 2\pi\sqrt{0.005}.

\sqrt{0.005} = 0.0707, so

T_\text{par} = 2\pi \times 0.0707 \approx 0.444 \text{ s}.

Why: parallel adds stiffness, so the combined spring is 400 N/m — stiffer than either spring alone. Stiffer means faster oscillation, shorter period.

Step 3 (series). Use the product-over-sum rule.

k_\text{eff}^\text{ser} = \frac{k_1 k_2}{k_1 + k_2} = \frac{100 \times 300}{100 + 300} = \frac{30{,}000}{400} = 75 \text{ N/m}.

Why: 1/k_\text{eff} = 1/100 + 1/300 = 3/300 + 1/300 = 4/300, so k_\text{eff} = 300/4 = 75 N/m. The series combination is softer than the softer of the two individual springs (75 < 100) — the chain is weaker than its weakest link.

Step 4 (series). Compute the period.

T_\text{ser} = 2\pi\sqrt{\frac{2}{75}} = 2\pi\sqrt{0.02667}.

\sqrt{0.02667} = 0.1633, so

T_\text{ser} = 2\pi \times 0.1633 \approx 1.026 \text{ s}.

Result: T_\text{par} \approx 0.44 s (parallel), T_\text{ser} \approx 1.03 s (series). The series configuration has a period about 2.3 times longer than the parallel one — because k_\text{eff}^\text{par}/k_\text{eff}^\text{ser} = 400/75 = 5.33, so the period ratio is \sqrt{5.33} \approx 2.31.

What this shows: Same two springs, same mass, different connection — wildly different periods. The lesson: always identify the connection style first. Parallel stiffens; series softens. And the difference is not a small correction; it can be a factor of two or more.

Example 3: Mass between two walls

A 0.5 kg block sits on a frictionless table between two fixed walls, connected to the left wall by a spring of stiffness k_1 = 200 N/m and to the right wall by a spring of stiffness k_2 = 50 N/m. Both springs are at their natural lengths when the block is at its equilibrium position.

(a) What is the time period of small oscillations about equilibrium? (b) If the block is displaced 2 cm to the right and released from rest, write down x(t) explicitly.

A 0.5 kg block between two walls. Spring $k_1 = 200$ N/m on the left, spring $k_2 = 50$ N/m on the right. Both are in parallel (they share displacement).

Step 1. Identify the effective stiffness.

Both springs stretch/compress by the same x when the block moves — they share displacement. Parallel rule:

k_\text{eff} = k_1 + k_2 = 200 + 50 = 250 \text{ N/m}.

Step 2. Compute \omega and T.

\omega = \sqrt{\frac{k_\text{eff}}{m}} = \sqrt{\frac{250}{0.5}} = \sqrt{500} \approx 22.36 \text{ rad/s}.

T = \frac{2\pi}{\omega} = \frac{2\pi}{22.36} \approx 0.281 \text{ s}.

Why: straight plug-in. Notice that even though the two springs have very different stiffnesses, only their sum appears — the block doesn't "see" them as two separate springs, it sees one effective spring of stiffness 250 N/m.

Step 3. Write down x(t).

Initial conditions: at t = 0, x_0 = 0.02 m and v_0 = 0 (released from rest). Using the cosine form is natural because the motion starts at the extreme:

x(t) = x_0 \cos(\omega t) = 0.02 \cos(22.36\, t) \;\text{m}.

Why: when the initial velocity is zero and the initial position is the extreme of the motion, A = x_0 and the phase works out so that x(t) = A\cos(\omega t). Differentiating confirms v(0) = -A\omega\sin(0) = 0, matching the initial condition.

Result: T \approx 0.28 s. x(t) = 0.02\cos(22.36\, t) m. The block oscillates between +2 cm and -2 cm with period 0.28 s.

What this shows: Asymmetric spring stiffnesses do not produce asymmetric motion. The two springs combine into one effective spring, and the oscillation is a symmetric SHM about the equilibrium position — provided both springs are Hookean (linear in displacement). The force contributions from the two springs are unequal (200 N/m versus 50 N/m), but they both oppose the displacement in the same direction, and their sum is what drives the motion.

Common confusions

"A vertical spring has a different period from a horizontal spring." False. The algebra looks different because gravity appears in the equilibrium calculation. But once you measure displacement from the new equilibrium, the equation of motion is identical to the horizontal case, and the period is the same T = 2\pi\sqrt{m/k}. Gravity shifts where the oscillation happens; it does not change how fast.
"Two springs between a mass and the wall are always in series." Not always. The question is not where they are drawn in the diagram — it is whether they share displacement or share force. Two springs anchored to the same wall and attached to the same mass share displacement (parallel). Two springs joined end-to-end in a chain share force (series). Two springs on opposite sides of a mass (each to its own wall) share displacement — both stretch/compress by the block's displacement — so also parallel.
"In the parallel case, the 'total force' means 2*kx, not 2k * x." Algebraically these are the same, but the conceptual distinction matters. Each spring contributes -k_1 x and -k_2 x independently; the sum of forces is -(k_1 + k_2)x. Do not imagine a single mysterious "combined spring force" that you have to derive — just add the individual forces.
"The period doesn't depend on amplitude, so a bigger push takes the same time." True for an ideal Hookean spring — this is the isochronism of SHM. In real life, very large amplitudes push the spring beyond its elastic limit, where Hooke's law fails and the period does depend on amplitude. Every formula in this chapter assumes you stay in the linear regime.
"Adding weight to a hanging spring changes its stiffness." No. Hanging a mass stretches the spring to a new equilibrium but does not change k; the spring is still equally responsive to extra displacement around its new equilibrium. The stiffness is a property of the spring, not of what you hang on it.
"If the spring has mass, the formula still works unchanged." Approximately, but not exactly. A realistic spring contributes a bit of its own mass to the effective inertia. The correction is covered in the going-deeper section: T = 2\pi\sqrt{(m + m_s/3)/k}, where m_s is the spring's mass. For laboratory springs this is usually a few-percent correction; for the ideal spring-mass problems in textbooks, assume the spring is massless.

If you have the period formula, the scaling T \propto \sqrt{m/k}, and the series/parallel rules, you have everything needed for most JEE problems on spring-mass systems. What follows adds three pieces of depth: the energy-conservation derivation (a second route to the period), the correction for a spring's own mass, and the two-mass (reduced-mass) problem.

The period from energy conservation

An alternative derivation uses conservation of energy instead of Newton's second law. For a horizontal spring-mass, the total mechanical energy is

E = \tfrac{1}{2} m v^2 + \tfrac{1}{2} k x^2,

with v = dx/dt. Because there is no friction or external force doing work, E is constant. Differentiate both sides with respect to time:

0 = \frac{dE}{dt} = m v \frac{dv}{dt} + k x \frac{dx}{dt} = m v a + k x v = v(m a + k x).

For the mass to oscillate, v is not identically zero. So the bracket must vanish:

m a + k x = 0 \quad\Rightarrow\quad m \frac{d^2 x}{dt^2} = -k x,

which is equation (1) again. This is often a cleaner derivation in complicated problems because it sidesteps the need to carefully list every force — you just write down the energy and differentiate.

Correction for the mass of the spring

In reality, a spring has its own mass m_s, distributed along its length. When the block is at position x, the spring is stretched to length L_0 + x; a point on the spring at a fraction \xi \in [0, 1] along its length (with \xi = 0 at the wall and \xi = 1 at the block) is at position \xi (L_0 + x), and its velocity is \xi \dot{x}. The kinetic energy of the spring is

KE_\text{spring} = \int_0^1 \tfrac{1}{2}\, (m_s\, d\xi)\, (\xi \dot{x})^2 = \tfrac{1}{2}\, m_s\, \dot{x}^2 \int_0^1 \xi^2\, d\xi = \tfrac{1}{2} \cdot \frac{m_s}{3}\, \dot{x}^2.

Why: slice the spring into infinitesimal segments of mass m_s\, d\xi. The segment at fraction \xi moves at speed \xi \dot{x} (linear interpolation — the wall end is fixed, the block end moves at \dot{x}). Each segment contributes \tfrac{1}{2}(m_s d\xi)(\xi \dot{x})^2. Integrate \xi^2 from 0 to 1 to get \tfrac{1}{3}.

So the kinetic energy of the system is

KE_\text{total} = \tfrac{1}{2} m \dot{x}^2 + \tfrac{1}{2} \cdot \frac{m_s}{3}\, \dot{x}^2 = \tfrac{1}{2}\left(m + \frac{m_s}{3}\right)\dot{x}^2.

The oscillation looks exactly like the massless-spring case but with an effective mass m_\text{eff} = m + m_s/3. The period becomes

\boxed{\; T = 2\pi\sqrt{\frac{m + m_s/3}{k}}. \;}

A laboratory spring might have m_s = 20 g and be used with a m = 200 g block; then m_\text{eff} = 206.7 g instead of 200 g — a 3% correction, affecting T by \sqrt{1.033} - 1 \approx 1.6\%. Small but measurable. For an Indian school-lab spring-constant experiment, this correction improves the agreement between the predicted and observed period from about 5% to about 1%.

Two masses on a spring — the reduced mass

Take two blocks of masses m_1 and m_2 connected by a spring of stiffness k, with nothing else attached (no walls, just the two blocks on a frictionless surface). Displace them symmetrically so the spring is stretched, and release. What is the period?

Let the positions be x_1 and x_2. The spring's extension is (x_2 - x_1) - L_0, where L_0 is the natural length. Define \ell = (x_2 - x_1) - L_0, the extension. The spring force on block 2 is -k\ell (toward block 1); on block 1 it is +k\ell (toward block 2) by Newton's third law. Newton's second law for each:

m_1 \ddot{x}_1 = k\ell, \qquad m_2 \ddot{x}_2 = -k\ell.

Subtract the first from the second:

\ddot{x}_2 - \ddot{x}_1 = -k\ell\left(\frac{1}{m_1} + \frac{1}{m_2}\right).

But \ddot{\ell} = \ddot{x}_2 - \ddot{x}_1 (since L_0 is constant), so

\ddot{\ell} = -k\left(\frac{m_1 + m_2}{m_1 m_2}\right)\ell = -\frac{k}{\mu}\,\ell,

where

\mu = \frac{m_1 m_2}{m_1 + m_2}

is the reduced mass of the two-body system. The equation for \ell is SHM with angular frequency \omega = \sqrt{k/\mu}, so

\boxed{\; T = 2\pi\sqrt{\frac{\mu}{k}} = 2\pi\sqrt{\frac{m_1 m_2}{k(m_1 + m_2)}}. \;}

Why: two free masses connected by a spring oscillate in a way that depends only on the relative separation, not on the centre-of-mass motion (which moves uniformly). The "effective mass" governing the relative oscillation is the reduced mass \mu — always smaller than either individual mass.

Quick check. Set m_2 \to \infty (one block much heavier — effectively a fixed wall). Then \mu = m_1 m_2 / (m_1 + m_2) \to m_1, and T \to 2\pi\sqrt{m_1/k} — the single spring-mass result. Perfect.

This reduced-mass trick reappears in every two-body oscillation problem — diatomic molecules vibrating in a crystal, binary stars, the hydrogen atom (with the tiny electron-proton mass ratio). For m_1 \ll m_2, \mu \approx m_1 and the heavier body acts as a wall. For m_1 = m_2 = m, \mu = m/2 — the system oscillates faster than if only one mass were present, because the effective inertia is halved.

An easy sanity check on series springs

Series springs can be derived in one line using the energy/compliance perspective. The compliance c = 1/k of a spring is the stretch per unit force. For springs in series, the same force produces extra stretch in each — compliances add:

c_\text{total} = c_1 + c_2 + \cdots \quad\Longleftrightarrow\quad \frac{1}{k_\text{total}} = \frac{1}{k_1} + \frac{1}{k_2} + \cdots.

This is the hydraulic-analogy way to see series springs: imagine the force as a "flow" and each spring's compliance as its "resistance to motion." Two compliances in series make twice as much give.

When does the small-oscillation formula break down?

All the formulas in this chapter assumed Hooke's law: F = -kx, exactly linear. Real springs follow Hooke's law only up to some maximum stretch (the elastic limit). Beyond that, the force grows more slowly or more quickly than linearly, the oscillation is no longer SHM, and the period starts to depend on amplitude. For the steel coil springs used in a car's suspension, the elastic limit is well beyond the usual operating range — you would have to hit the pothole at 120 km/h to leave the linear regime. For a stretched rubber band, the nonlinearity sets in much sooner, which is why rubber bands don't give perfect SHM.

Where this leads next

Energy in SHM — how kinetic energy and spring potential energy trade off during the oscillation, always summing to the constant E = \tfrac{1}{2} k A^2.
Simple Pendulum — a different realisation of SHM, with T = 2\pi\sqrt{L/g} in place of T = 2\pi\sqrt{m/k}. Same equation, different physical constants.
Damped Oscillations — add a velocity-dependent damping force (like a shock absorber on the Alto's suspension) and the amplitude shrinks exponentially while the period changes slightly.
Forced Oscillations and Resonance — drive the spring-mass with a periodic external force, and watch the system oscillate at the driving frequency — with the amplitude peaking sharply when the driving frequency matches the natural frequency \omega_0 = \sqrt{k/m}.
Spring Force and Hooke's Law — the underlying force law that makes all of this possible, and the limits on its validity.