Edexcel IAL - Mechanics 3- 3.2 Simple Harmonic Motion- Study notes - New syllabus
Edexcel IAL – Mechanics 3- 3.2 Simple Harmonic Motion -Study notes- New syllabus
Edexcel IAL – Mechanics 3- 3.2 Simple Harmonic Motion -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 3.2 Simple Harmonic Motion
Simple Harmonic Motion
A particle is said to move with simple harmonic motion (SHM) if its acceleration is always directed towards a fixed point and is proportional to its displacement from that point.
Defining Equation of SHM
If the displacement of a particle from its equilibrium position is \( x \), then SHM is defined by the equation
\( \ddot{x} = -\omega^2 x \)
where \( \omega \) is a positive constant called the angular speed.
The negative sign indicates that the acceleration is always directed towards the equilibrium position.
Showing That Motion is SHM
To prove that a particle executes SHM in a given situation, it is sufficient to show that its acceleration satisfies
Acceleration \( \propto – \) displacement
That is, the equation of motion can be written in the form
\( \ddot{x} = -\omega^2 x \)
This may be done using either Newton’s laws, energy methods, or geometrical arguments.
Standard Formulae for SHM
For a particle executing SHM with angular speed \( \omega \):
Displacement: \( x = A\cos(\omega t) \) or \( x = A\sin(\omega t) \)
Velocity: \( v = -A\omega\sin(\omega t) \) or \( v = A\omega\cos(\omega t) \)
Acceleration: \( a = -\omega^2 x \)
The period of the motion is
\( T = \dfrac{2\pi}{\omega} \)
Energy in SHM
The total mechanical energy of a particle executing SHM is constant and equal to
\( \dfrac{1}{2}m\omega^2A^2 \)
This energy continually transforms between kinetic and potential energy.
Example :
A particle moves along a straight line such that its acceleration is given by \( a = -9x \). Show that the motion is simple harmonic and find the angular speed.
▶️ Answer/Explanation
The given equation of motion is
\( \ddot{x} = -9x \)
Comparing with \( \ddot{x} = -\omega^2 x \), we identify
\( \omega^2 = 9 \Rightarrow \omega = 3 \)
Conclusion: The motion is SHM with angular speed \( 3 \,\text{rad s}^{-1} \).
Example :
A particle of mass \( 0.5 \) kg is attached to a horizontal spring of spring constant \( 8 \,\text{N m}^{-1} \). Show that the particle performs SHM and find its period.
▶️ Answer/Explanation
For a spring, the restoring force is
\( F = -kx \)
Using Newton’s second law:
\( m\ddot{x} = -kx \)
\( \ddot{x} = -\dfrac{k}{m}x \)
So
\( \omega^2 = \dfrac{8}{0.5} = 16 \Rightarrow \omega = 4 \)
The period is
\( T = \dfrac{2\pi}{4} = \dfrac{\pi}{2} \)
Conclusion: The motion is SHM with period \( \dfrac{\pi}{2} \) s.
Example :
A particle executes SHM with angular speed \( 5 \,\text{rad s}^{-1} \) and amplitude \( 0.2 \) m. Find its maximum speed and maximum acceleration.
▶️ Answer/Explanation
Maximum speed in SHM is
\( v_{\max} = \omega A \)
\( v_{\max} = 5 \times 0.2 = 1 \)
Maximum acceleration is
\( a_{\max} = \omega^2 A \)
\( a_{\max} = 25 \times 0.2 = 5 \)
Conclusion: The maximum speed is \( 1 \,\text{m s}^{-1} \) and the maximum acceleration is \( 5 \,\text{m s}^{-2} \).