Oscillations · Lesson 4 of 8
Spring Oscillators and Their Applications
“Build the restoring-force equation for spring systems and solve spring-balance, paired-spring and two-free-mass applications.”
• Derive the single-spring period with the equilibrium and model assumptions stated. • Find the effective stiffness of two fixed opposed springs. • Calibrate a spring balance and infer mass and weight from an oscillation period. • Distinguish extension under equal end forces from the displacement of each free end. • Calculate frequency, maximum acceleration and maximum speed with consistent units.
Why a spring returns a displaced block
Attach a block of mass m to a light spring on a smooth horizontal surface. Within its elastic range, the spring force is proportional to its extension or compression. When the spring’s unstretched position is also the block’s horizontal equilibrium, a displacement x to the right produces F = −kx. For x < 0 the compressed spring pushes right. Vertical weight and the normal reaction balance, so they do not enter the horizontal equation. Neglect spring mass, friction and other horizontal forces; k is treated as constant.
A stiffer spring produces a larger restoring acceleration for the same displacement, so its period is shorter. A larger mass has more inertia, so its period is longer. Doubling amplitude changes maximum force and speed, but it does not change the ideal period: both displacement and restoring force grow together. These amplitude-independent results require the linear approximation. They cannot be transferred unchanged to a spring stretched beyond its Hooke-law range.
Two fixed springs: count both forces
Problem
A block m lies on a smooth surface between identical light springs of stiffness k, with the outer ends fixed. Show that its displacement from equilibrium produces SHM and find T.
- 1.Displace the block a distance x to the right. The left spring lengthens by x and pulls the block left: F₁ = −kx.
- 2.The right spring shortens by x and pushes the block left: F₂ = −kx. Compression does not reverse the restoring direction on this block.
- 3.Add the forces on the same body: Fnet = −2kx. The same equation also restores a leftward displacement, for which both forces are rightward.
- 4.Newton’s law gives x″ = −(2k/m)x. Therefore ω = √(2k/m) and T = 2π√[m/(2k)]. The two springs act with effective stiffness 2k, not k/2.
For two opposed fixed springs with constants k₁ and k₂, the force changes relative to equilibrium are −k₁x and −k₂x, so the effective stiffness is k₁ + k₂. Any pre-existing constant forces must already balance at the chosen equilibrium. This argument adds forces acting on one displaced mass; it does not establish a universal rule for every arrangement containing two springs. Identify how each spring’s length changes before adding forces.
A spring balance converts a static calibration into a dynamic measurement
In a vertical spring, gravity creates a static equilibrium extension y₀ with ky₀ = mg. Let downward y measure extension from the natural length, and let x = y − y₀ measure displacement from equilibrium. Then m x″ = mg − k(y₀ + x) = −kx. Gravity shifts the equilibrium but cancels from the equation for x. Consequently the small vertical oscillation period is again 2π√(m/k), provided the spring and support obey the model. Do not substitute the full static extension y for the equilibrium displacement x in the restoring equation.
Problem
A balance’s 0–50 kg calibration occupies a length of 20 cm. A suspended body oscillates with period 0.6 s. Find its weight using g = 9.8 m s⁻², neglecting the moving mass of the spring and pan.
- 1.The scale’s “50 kg” mark means the force due to a 50 kg mass under the calibration gravity: F = 50 × 9.8 = 490 N. The corresponding extension is 0.20 m.
- 2.Thus k = 490/0.20 = 2450 N m⁻¹. Treat the calibration as linear.
- 3.Rearrange T² = 4π²m/k to m = kT²/(4π²). Substitution gives m = 2450 × 0.36/(4π²) = 22.3413 kg.
- 4.The weight is mg = 218.945 N, approximately 219 N. Reporting 22.3 N would confuse mass with weight; reporting 50 kg would use the scale maximum instead of the oscillating load.
One data set determines frequency, speed and acceleration
Problem
A horizontal 3 kg block attached to a 1200 N m⁻¹ spring is pulled 2.0 cm from equilibrium and released from rest. Find its frequency, maximum acceleration and maximum speed.
- 1.Release from rest at an extreme means A = 2.0 cm = 0.020 m. The angular frequency is ω = √(1200/3) = 20 rad s⁻¹.
- 2.The frequency is f = ω/(2π) = 3.18310 Hz, approximately 3.2 Hz. Its period is π/10 s ≈ 0.314 s.
- 3.The maximum acceleration magnitude is ω²A = 400 × 0.020 = 8.0 m s⁻², reached at either extreme.
- 4.The maximum speed is ωA = 20 × 0.020 = 0.40 m s⁻¹, reached at equilibrium.
- 5.At the release point v = 0 despite the maximum acceleration magnitude. Changing when the stopwatch starts changes only phase; the same oscillator’s alternative equations were derived in the phase lesson.
Problem
A piston has a full extreme-to-extreme stroke of 1.0 m and is modelled as SHM with angular frequency 200 rad min⁻¹. Find its greatest speed.
- 1.The amplitude is half the stroke: A = 0.50 m.
- 2.Use vmax = ωA = (200 rad min⁻¹)(0.50 m) = 100 m min⁻¹.
- 3.In SI seconds this is 100/60 = 1.66667 m s⁻¹. Alternatively convert ω to 200/60 rad s⁻¹ before multiplying.
- 4.Do not multiply by 2π: the given rate is already an angular frequency, not a cycle frequency. Real crank-driven pistons need not be exact SHM; the calculation uses the stipulated model.
One spring with two free masses is a different system
First distinguish loading from free oscillation. A light spring held with a force of magnitude F at each end has tension F, not 2F: either end transmits the same tension through the spring. Its static total extension is q₀ = F/k. With one end clamped and one pulled quasistatically by F, the static extension is also F/k. “Maximum extension” in this loading comparison means the initially held extension, reached quasistatically with no kinetic energy; an unspecified sudden loading history could produce a different transient maximum.
After both equal masses m are released from rest, symmetry keeps the midpoint fixed. Let u be the right mass’s outward displacement from its position at natural spring length; the left mass’s outward displacement has equal magnitude. The total spring extension is q = 2u. Its tension is kq = 2ku, which pulls the right mass inward. Thus m u″ = −2ku, so each mass has angular frequency √(2k/m) and period 2π√[m/(2k)]. The extension q oscillates with the same period but twice the displacement amplitude of either mass. For the clamped version only one mass moves and T = 2π√(m/k).
This result happens to equal the period of one mass between two equal fixed springs, but the systems are physically different. In the fixed-support case each spring changes length by x. In the free-mass case one spring changes length by 2u. The equations agree only after those different coordinates and constraints are identified. For a given held tension F, each free mass starts only F/(2k) from its equilibrium position, while the whole spring’s extension is F/k.
Quick Quiz
Quiz
A single ideal spring’s k increases by a factor 4 with m fixed. How does T change?
T ∝ 1/√k, so multiplying k by 4 divides the period by 2.
For a block between equal fixed springs, a right displacement produces what net force?
The stretched left spring and compressed right spring both push or pull the block left.
A light spring is held by equal opposite end forces of magnitude F. Its tension is
The same transmitted force balances either end. Adding external vector forces is not the way to calculate internal tension.
Two equal free masses joined by one spring each move outward by u. How much does the spring length increase?
One end moves left by u and the other moves right by u, so their separation increases by 2u.
A piston’s stroke is 0.8 m and ω = 10 rad s⁻¹. What is vmax?
A = 0.4 m, so vmax = ωA = 4 m s⁻¹.
Practice Problems
- Derive the period of a 0.5 kg mass on a 200 N m⁻¹ horizontal spring. Explain why weight does not enter its horizontal motion equation.
- A block of mass 2 kg lies between two opposed fixed springs of constants 100 N m⁻¹ and 300 N m⁻¹. Derive the net restoring force, angular frequency and period.
- A spring balance extends 0.10 m for a 20 kg calibration mass. An unknown load oscillates with T = 0.50 s. Using g = 9.8 m s⁻², find k, the load’s mass and its weight, neglecting spring and pan mass.
- For m = 3 kg, k = 1200 N m⁻¹ and release from rest 2.0 cm from equilibrium, compute f, T, vmax and |a|max. State where each maximum occurs.
- Compare a spring pulled quasistatically by F with its other end clamped and a spring pulled by equal opposite forces F with an equal mass m on each free end. Find held total extension and the free-oscillation period in each case. Explain why the free case does not have tension 2F.
- A piston has stroke 1.0 m and angular frequency 200 rad min⁻¹. Calculate vmax in both m min⁻¹ and m s⁻¹, explaining both conversions.
- Show algebraically why a vertical ideal spring has the same period as its horizontal version when displacement is measured from the gravity-shifted equilibrium. Distinguish static extension from oscillatory displacement.
Key Takeaways
• A constant linear restoring coefficient gives ω = √(k/m) and T = 2π√(m/k). • Identify how each spring length changes before adding restoring forces. • Equal opposite end forces F give spring tension F, not 2F. • In a two-free-equal-mass system the extension is twice either mass’s displacement about the fixed midpoint. • Spring balances measure force through calibration; mass and weight must be kept distinct.