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When one end of a spring is fixed to a wall, a block on it oscillates with T = 2π√(m/k). Join two free blocks with a spring and set them moving, and neither end is fixed. Which mass goes into the formula? The answer is the reduced mass, m₁m₂/(m₁ + m₂). This post derives it two ways, works two problems with checks, and lists the traps.
Is this in the syllabus?
NCERT's chapter Oscillations treats a block on a spring. The chapter Systems of Particles and Rotational Motion shows in section 6.4 that when the total external force is zero, the velocity of the centre of mass stays constant. Two-body spring problems combine the two. The JEE (Advanced) 2026 Information Brochure lists "Centre of mass and its motion" and "Linear and angular simple harmonic motions", and the Syllabus for JEE (Main) 2026 lists "oscillations of a spring: restoring force and force constant". Check the current year's syllabus before your exam.
Where does the reduced mass come from?
Put the blocks on a smooth floor at positions x₁ and x₂, joined by a spring of natural length L₀. Let x = x₂ − x₁ − L₀ be the extension. The spring pulls them together with force kx:
m₁ × (acceleration of block 1) = +kx
m₂ × (acceleration of block 2) = −kx
The extension's acceleration is the second minus the first:
acceleration of x = −k(1/m₁ + 1/m₂)x = −(k/μ)x, where μ = m₁m₂/(m₁ + m₂)
That is SHM in the extension, with ω = √(k/μ) and T = 2π√(μ/k).
Is there a quicker way to see it?
Yes. The centre of mass does not accelerate, so stand on it. The point of the spring at the centre of mass stays still, splitting the spring into two pieces. The piece attached to m₁ is a fraction m₂/(m₁ + m₂) of the whole, and a shorter spring is stiffer, so its constant is k(m₁ + m₂)/m₂. Then ω² = k(m₁ + m₂)/(m₁m₂) = k/μ, the same answer.
Worked example 1: released from rest
Blocks of 1.0 kg and 3.0 kg on a smooth floor are joined by a spring of constant 300 N/m. The spring is compressed by 4.0 cm and released.
μ = (1.0 × 3.0)/(1.0 + 3.0) = 0.75 kg.
ω = √(300/0.75) = √400 = 20 rad/s, so T = 2π/20 ≈ 0.314 s.
Total momentum is zero, so the centre of mass stays put. The extension swings between −4.0 cm and +4.0 cm.
The blocks share the motion in inverse ratio to mass: the 1.0 kg block moves 3.0 cm either side of its mean position, the 3.0 kg block 1.0 cm.
Maximum relative speed = ωA = 20 × 0.040 = 0.80 m/s, shared as 0.60 m/s and 0.20 m/s.
Check momentum: 1.0 × 0.60 = 3.0 × 0.20. Check energy: ½ × 300 × 0.040² = 0.24 J, and ½ × 1.0 × 0.60² + ½ × 3.0 × 0.20² = 0.18 + 0.06 = 0.24 J.
Worked example 2: one block is struck
Two 2.0 kg blocks rest on a smooth floor, joined by a spring of constant 400 N/m at its natural length. One block is suddenly given 1.0 m/s toward the other. Find the maximum compression.
The centre of mass moves at (2.0 × 1.0)/4.0 = 0.50 m/s throughout, carrying ½ × 4.0 × 0.50² = 0.50 J that the spring can never take.
At maximum compression both blocks move at 0.50 m/s, so the spring holds 1.0 − 0.50 = 0.50 J. Equivalently, ½μv_rel² = ½ × 1.0 × 1.0² = 0.50 J.
½ × 400 × x² = 0.50 gives x = 0.050 m = 5.0 cm.
With μ = 1.0 kg, ω = 20 rad/s, so maximum compression comes a quarter period later, after about 0.079 s.
Setting ½kx² equal to the full 1.0 J gives about 7.1 cm, the most common wrong answer.
Which mistakes cost marks?
Using m₁ + m₂, or one of the masses, in the period.
Giving the spring all the kinetic energy and forgetting that the centre of mass keeps moving.
Thinking the heavier block stays still. It moves less, not zero.
Mixing up the amplitude of the extension with each block's own amplitude.
The same reduced mass appears in gravitation, where two stars orbit their common centre of mass; see binary stars and two-body orbits. For single-spring energy problems, see spring-block systems.
How should you practise this?
Chi, Feltovich and Glaser (1981) found that physics experts start from the principle that solves a problem, while novices start from its surface features. Two-body problems reward the expert habit: ask first what is conserved (momentum always, mechanical energy if there is no friction) and only then what oscillates.
Day 1: derive ω = √(k/μ) both ways from a blank page.
Day 2: three release-from-rest problems with different mass ratios.
Day 3: two struck-block problems, finding the maximum compression and the time to reach it.
Day 4: a mixed set that includes a wall-fixed spring, so you must decide which case you have.
When you cannot tell what is conserved, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, redo example 2 with a struck 1.0 kg block and a 3.0 kg partner, and check that your momentum and energy lines still balance.
