S6.4 The SHM equation: applications
[A] The task and the strategy
- We have established (S6.2)
that if a system in stable equilibrium is given a displacement x
the resulting time-evolution of x will follow the SHM equation
where ω follows from the properties of the system.
- We have also established that the general solution to this
equation of motion is
x(t)=xmcos(ωt+φ)where xm and φ follow from the initial conditions.
- To apply the general theory to any specific case we are faced with one task: we must determine the SHM frequency.
- The strategy is always the same:
- We identify an arrangement of stable mechanical equilibrium, whose signature is that it is a minimum of potential energy.
- We think of what happens if the system is displaced by a small amount x from equilibrium; we usually have a choice as to what x will represent.
- We use Newton’s laws to set up the equation of motion for x.
- If all goes well we find that this equation can be written
in the form
where the something-or-other is determined in terms of the properties of the system.
- We conclude that the x-coordinate of this system
will exhibit SHM with angular frequency
- This concludes the job: anything else we need is already available in our general SHM results.
- The following applications show this strategy at work.
[B] Application: mass on spring
A mass m rests on a horizontal frictionless surface, and is attached to a wall by a spring of force constant k. Establish the motion if the mass is pulled a small distance (stretching the spring from its natural length), and then released. [Yes: you have seen this example before; it is useful to help establish the sequence of steps in the strategy; you should work through it yourself first; then check what you have done against the solution that follows .] |
Solution
Reveal

Solution
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- Choose a system to which to apply Newton’s 2nd Law: the mass
- Identify the force(s) acting on this system:
F=-kx
- Write down the equation of motion:
- Rearrange this equation in the standard form and identify ω:
Results
Key Point 6.4
A mass m subject to a linear spring-force of spring-constant k will exhibit SHM of angular frequency![\[ \omega= \sqrt{\frac{k}{m}} \]](mastermathpng-5.png)
- Are the units OK?
- Does it make sense?
[C] Application: hydrogen molecule
A molecule of hydrogen can be modelled as two particles of mass mH=1.7×10-27 kg linked by a bond viewed as a spring of spring constant k=5.2×102 N⋅m-1 Establish the motion if the bond is compressed and then released. |

Solution
- Identify stable equilibrium: the spring (’bond’ is unstretched)
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- Choose the coordinate x: a displacement of (say) mass 2 to to the right.
- Choose a system to which to apply Newton’s 2nd Law: the mass 2
- Identify the force(s) acting on this system:
The spring length is changed by 2x
since both masses move a distance x in opposite directions
- The force on mass 2 is then
F=-2kx
Write down the equation of motion:
Rearrange this equation in the standard form and identify ω:
- Insert numbers
(…but not before now: recall
Guideline 0.6)
So the frequency is
Results
The molecular bond (spring) exhibits SHM with angular frequency
![\[ \omega= \sqrt{\frac{2k}{m}} =7.8 \times 10 ^{14} rad \cdot s^{-1} \]](mastermathpng-11.png)

Commentary
- This frequency lies in the near-infrared part of the electromagnetic spectrum
- The bond-stretching-frequencies of other molecules are generally lower because ’other’ atoms are heavier than hydrogen!
- The frequency provides a distinctive signature of the molecule –apparent in the spectrum of light emitted by or absorbed by gases containing such molecules.
[D] Application: simple pendulum
A simple pendulum comprises a point mass m suspended from a fixed point by a string of length L. Establish the motion if the bob is pulled to one side and then released. |

Solution
- Identify stable equilibrium: the bob hangs vertically
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Help? Don’t miss this one!
The approximations for sinθ and cosθ for small θ are amongst the most important examples of Taylor series.- So the vertical displacement will be negligible compared with the horizontal displacement,
- Choose the coordinates: x (horizontal) and y (vertical).
- Choose a system to which to apply Newton’s 2nd Law: the bob
- Identify the forces acting on this system
- The tension T along the string
- The gravitational force mg downwards
- Write down the equation of motion:
- In the x-direction:
- In the y-direction:
- If y is ’small’, then so is
so the y-equation implies
T≃mg Substitute into the x equation:
where the last step uses the fact that
for small θ using the Taylor expansions for sin and cos again.
- In the x-direction:
- Rearrange the x equation in the standard form and identify ω:
Results
Key Point 6.5
A pendulum comprising a point mass suspended from a fixed point by a light string of length L exhibits SHM of angular frequency![\[ \omega= \sqrt{\frac{g}{L}} \]](mastermathpng-19.png)
- Are the units OK?
- Does it make sense?
[E] Application: physical pendulum
A physical pendulum is made from a sheet of steel, of mass m, pivoted about an axis through a point a distance d from its centre of mass. The moment of inertia about this axis is I. Establish the motion if the sheet is pushed to one side and then released. |

Solution
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Comment
Remember that in the SHM equation the oscillating quantity doesn’t have to be a displacement; this is the first example we have met where it isn’t. While we could choose to call the angle ’x’ if we really wanted, it is better to use a symbol that reminds us what kind of quantity we are dealing with.- Choose a system to which to apply Newton’s 2nd Law:the plate
- Identify the forces acting on this system:
- The weight of the plate acting downwards, through the CoM
- The reaction force exerted by the pivot acting through the pivot
- Write down the equation of motion:
- We need the angular form
of Newton’s 2nd Law
(because we have chosen to consider an angular coordinate!
(Key Point 5.19)
where I is the moment of inertia about the pivot axis, and τ is the torque about this axis.
- The torque due to the weight has magnitude
τ=mgdsinθ
The equation of motion is (after some thought about signs)
Do you want to understand the minus sign?
You do? Good!
First note that if we missed this minus sign we would rapidly find that we had done something wrong. Instead of the standard SHM equation we would find that θ had the equation of motion
This equation gives an altogether different behaviour: θ would go on growing at an ever greater rate!This physical nonsense comes about if we fail to capture the fact that the torque has a sense (direction): it is a restoring torque that always acts back towards equilibrium. The direction of a torque is expressed through its vector character, which is not apparent in the equation τ=Iα. In this equation the LHS gives the component of the torque vector along the chosen axis; the RHS gives the rate of change of the angular momentum along the same axis.
The component of the torque along the axis into the plane is:
τ=mgdsinθie it is into the plane (positive) if θ is positive, and out of the plane (negative) if θ is negative.The angular momentum along the axis into the plane is:
ie it is out of the plane (negative) ifis positive, and into the plane (postive) if
is negative.
Then the equation
does indeed implyPhew!- Now suppose that θ is small so we can
use the Taylor Series expansion
for sinθ:
- We need the angular form
of Newton’s 2nd Law
(because we have chosen to consider an angular coordinate!
(Key Point 5.19)
- Rearrange the θ equation in the standard form and identify
ω:
Results
The physical pendulum exhibits SHM of angular frequency
provided the angular displacement from the vertical is small.
[F] The core physics of SHM: stiffness versus inertia
- We can identify a common underlying structure in these examples.
Key Point 6.6
The SHM frequency ω is determined by the properties of the system; it can be written in the generic form - By ’stiffness’ we mean strength of effects restoring equilibrium (force, torque, …)
- By ’inertia’ we mean strength of effects resisting changes in motion (mass, moment of inertia …)
- Oscillation results from an interplay between the two: ’stiffness’ favours return to equilibrium; inertia results in overshoot beyond equilibrium.
- While ’stiffness’ and ’inertia’ may each take different forms, if correctly identified their ratio has units 1/[second]2
- These insights are enough to allow one to estimate frequencies for ’new’ problems...with virtually no mathematics!

Example
Star oscillations
- A typical star exhibits oscillations perpendicular to the galactic plane in addition to its motion across the plane.
- These oscillations are roughly simple harmonic
- The parameters controlling the frequency might be
- the galactic density ρ ?
- the gravitational constant G ?
- the star mass M ?
- A guess at the stiffness:
stiffness∼GρM ?
- A guess at the inertia:
inertia∼M?
- The guesstimate for the frequency:
- The units check:
- A full analysis gives:
- Setting G≃6×10-11 and ρ≃10-18 (in SI units) leads to a period of order 108 years.
Learning Resources
![]() | HRW 15.5-15.6 |
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