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Theme C — Wave Behaviour

C.1.3 Time Period of a Mass–Spring System — IB Physics HL Revision Notes & Topical Questions

5 min read 10 parts Lesson 5 of 37 1 quiz

📚 IB Physics HL Hub / Topical Revision
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Lesson summary

C.1.3 Time Period of a Mass–Spring System — IB PHYSICS SL notes: Use T = 2π√(m/k) to find the period of a mass on a spring and explain what affects it.System — a mass m on a light spring of spring constant k oscillates horizontally on a smooth surface or hangs verticallyHooke's l…

  • Hooke's law
  • Newton's second law
  • Comparison
  • Rearranged form
  • Both orientations
  • Equilibrium

Part 1 of 8The mass–spring system

The mass–spring system
System
  • System a mass m on a light spring of spring constant k oscillates horizontally on a smooth surface or hangs vertically
Hooke's law
  • Hooke's law the spring force is F = −kx, where x is the extension or compression from the natural length
  • k is in newtons per metre (N m−1).
  • The minus sign shows the force opposes the displacement.
SHM
  • F = −kx is the restoring force that makes the motion simple harmonic (C.1.2)

Part 2 of 8Deriving the period

Newton's second law
  • Newton's second law ma = −kx, so a = −(k/m)x
Comparison
  • Comparison with a = −ω2x this gives ω2 = k/m
Period
  • T = 2π√(m/k), where T is in s, m in kg and k in N m−1
Frequency
  • Frequency f = (1/2π)√(k/m)
Rearranged form
  • Rearranged form k = 4π2mf2
Both orientations
  • Both orientations the equation applies to horizontal and vertical systems

Part 3 of 8What affects the period

Mass
  • T ∝ √m, so four times the mass doubles the period
  • A heavier object is harder to accelerate.
Spring constant
  • T ∝ 1/√k, so a stiffer spring gives a shorter period
  • A larger k gives a larger restoring force.
Amplitude
  • Amplitude it has no effect while Hooke's law holds
  • The spring must not be stretched beyond its elastic limit.
Gravity
  • Gravity the period is independent of g, so the same system oscillates with the same period on Earth and on the Moon
  • Gravity changes the equilibrium position, not the period.

Part 4 of 8Vertical springs

Equilibrium
  • Equilibrium the weight is balanced by the spring force, mg = ke, where e is the static extension
Motion about it
  • Motion about it displacing the mass by x from the new equilibrium gives net force −kx, so it is SHM with the same formula
Static extension
  • Static extension k/m = g/e, so T = 2π√(e/g)
Tension
  • Tension the spring must stay under tension throughout, because a slack spring cannot supply a restoring force

Part 5 of 8Experiments with a spring

1
Hang different masses from the spring and measure the extension for each.
2
Plot weight against extension; the gradient is k.
3
For several masses, time 10 or more oscillations and divide to find T.
4
Plot T2 against m.
Result
  • T2 = (4π2/k)m gives a straight line through the origin with gradient 4π2/k
Timing
  • Timing timing 10 or more oscillations reduces the percentage uncertainty due to reaction time
Limit
  • Limit if a force–extension graph curves at large loads, the limit of proportionality is passed and the SHM equations no longer apply

Part 6 of 8Springs in combination

Springs in combination
Two identical springs in parallel
  • Two identical springs in parallel they share the load, so the effective constant is 2k
  • The period is shorter by a factor of √2.
Two identical springs in series
  • Two identical springs in series the effective constant is k/2
  • The period is longer by a factor of √2.
Trolley between two springs
  • Trolley between two springs both springs act on it, so the effective constant is 2k

Part 7 of 8Assumptions and refinements

Light spring
  • Light spring the formula assumes a spring of negligible mass and negligible friction
Spring with mass
  • Spring with mass the effective oscillating mass is about m + ms/3
Damping
  • Damping air resistance makes the amplitude decay but leaves the period almost unchanged
Natural frequency
  • Natural frequency the frequency from f = (1/2π)√(k/m) is the natural frequency of the system, used in resonance

Part 8 of 8More from the specification

More from the specification
Period of Mass-Spring Oscillator and Pendulum
  • The time period of a mass-spring system is given by T = 2 pi sqrt(m/k), dependent on mass and spring constant,
  • whereas for a simple pendulum at small angles, period is T = 2 pi sqrt(L/g).
    • (theta < 10 degrees)
Conservation of Energy in Harmonic Systems
  • In simple harmonic motion, total mechanical energy is conserved;
    • kinetic energy reaches a maximum at zero displacement (equilibrium),
  • while potential energy is maximum at peak displacements (+/- x0).
Velocity and Acceleration in SHM
  • Maximum velocity occurs as the oscillator passes through equilibrium; acceleration is maximum at extreme displacements and directed towards equilibrium with a negative sign (a = -omega2 x).
    • (vmax = omega x0)
Energy-Displacement Graphs
  • On an energy-displacement graph, total mechanical energy is a horizontal line, kinetic energy forms an inverted parabola with peak at x = 0,
  • and potential energy forms an upright parabola touching zero at equilibrium.
Kinetic energy
  • K = ½mv2, energy of motion.
Potential energy
  • Energy stored by the configuration of a system.
Angular frequency
  • Ω = 2π/T = √(k/m) for a spring.
Gravitational potential
  • Work done per unit mass in bringing a test mass from infinity to the point.
Displacement-time graph
  • A graph of displacement against time whose gradient is velocity.
Angular displacement
  • Angle turned, in radians.
Velocity-time graph
  • A graph of velocity against time whose gradient is acceleration and area is displacement.
Pendulum
  • A weight swinging on a string, exchanging GPE and KE.
Velocity
  • Rate of change of displacement.

Key terms

TermMeaning
Spring constant kForce per unit extension, in N m⁻¹.
Hooke's lawForce is proportional to extension within the elastic limit.
Restoring forceF = −kx directed to the equilibrium position.
Static extensionExtension of a vertical spring when the mass is at rest.
Elastic limitThe extension beyond which the spring does not return to its original length.

Avoid these mistakes

1
Using mg alone as the spring constant — k = F/x needs the extension in metres.
2
Thinking the amplitude affects the period.
3
Saying a different value of g changes the period of a mass–spring system — it does not.
4
Writing T = 2π√(k/m) with the ratio inverted.
5
Using centimetres for the extension without converting to metres.

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