Further Mathematics | Further kinematics

Welcome to Further Kinematics (M3)!

Hello future mechanical engineers and physicists! You've successfully navigated constant acceleration (SUVAT) and even tackled variable acceleration using differentiation and integration in M2. Now, in M3, we go deeper.

This chapter is vital because it introduces you to the mathematics of oscillating systems—motion that repeats itself, which is everywhere in the real world (from sound waves to pendulums). Our main focus here is Simple Harmonic Motion (SHM).

Don't worry if this seems tricky at first; we will break down the calculus and connect it to forces and energy, making it much clearer. Let’s dive in!

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Section 1: The Foundation – Variable Acceleration Review

In Further Kinematics, motion is rarely constant. Acceleration often depends on time, distance, or velocity. To solve these problems, we rely heavily on calculus.

1.1 The Kinematic Chain (Calculus Toolkit)

The position of a particle is usually denoted by \(x\) (displacement) or \(s\). Time is \(t\).

Going Down the Chain (Differentiation)

If you have displacement and want acceleration, you differentiate twice:

• Velocity: \(v = \frac{dx}{dt}\)
• Acceleration: \(a = \frac{dv}{dt} = \frac{d^2x}{dt^2}\)

Going Up the Chain (Integration)

If you have acceleration and want displacement, you integrate twice. Crucially, remember the constant of integration, \(C\), which requires initial conditions (like speed at \(t=0\)).

• Velocity: \(v = \int a \, dt\)
• Displacement: \(x = \int v \, dt\)

The v vs x Relationship (The M3 Favourite)

When acceleration is given as a function of displacement, \(a = f(x)\) (which happens often in SHM!), we need a special formula to find velocity:

$$a = v \frac{dv}{dx}$$

Why is this tricky? When using this formula, you must separate the variables before integrating. For example, if \(a=4x\):

\(\int v \, dv = \int 4x \, dx\)

Key Takeaway: M3 Kinematics is built on the calculus foundation from M2. Master the relationship \(a = v \frac{dv}{dx}\).

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Section 2: Simple Harmonic Motion (SHM)

This is the central topic of Further Kinematics. SHM describes the oscillation of a particle about a fixed point.

2.1 Defining Simple Harmonic Motion

SHM occurs when two conditions are met:

1. The acceleration (\(a\)) is always directed towards a fixed point (the equilibrium centre, usually \(O\)).
2. The magnitude of the acceleration is proportional to the distance (\(x\)) from that fixed point.

This definition translates into the fundamental differential equation for SHM:

$$a = - \omega^2 x$$

Wait, what is \(\omega\)? \(\omega\) (omega) is a constant known as the angular frequency. It dictates how fast the system oscillates. Since \(a\) must be positive when \(x\) is negative, and negative when \(x\) is positive, we must have the negative sign in the equation.

Analogy: The Bouncing Spring
Imagine a mass attached to a spring oscillating horizontally. When you pull the mass far to the right (\(x\) is large and positive), the spring pulls it back hard (acceleration \(a\) is large and negative). When the mass is far left (\(x\) is large and negative), the spring pushes it hard right (acceleration \(a\) is large and positive). This is SHM!

2.2 The General Solutions for SHM

If \(a = -\omega^2 x\), we can solve this second-order differential equation to find the displacement, \(x\), as a function of time, \(t\).

The general solution is a wave function (sine or cosine):

$$x = A \cos(\omega t) \quad \text{or} \quad x = A \sin(\omega t)$$

(We choose which function based on the initial conditions at \(t=0\)).

• \(A\) is the Amplitude: The maximum displacement from the centre \(O\).
• \(\omega\) is the Angular Frequency: Related to the speed of oscillation.

Step-by-Step Example (Finding Velocity)
If \(x = A \cos(\omega t)\), we differentiate to find \(v\):

$$v = \frac{dx}{dt} = -A\omega \sin(\omega t)$$

Did you know? If you differentiate \(v\) again, you get \(a = -A\omega^2 \cos(\omega t)\). Since \(A \cos(\omega t) = x\), we confirm that \(a = -\omega^2 x\). The physics works!

2.3 Key Parameters of SHM

Once you find \(\omega\), you can calculate the period and frequency:

I. Period (\(T\))

The time taken for one complete oscillation (e.g., going from maximum displacement right, back through the centre, to maximum left, and returning to the start).

$$T = \frac{2\pi}{\omega}$$

Units: Seconds (s).

II. Frequency (\(f\))

The number of complete oscillations per second. It is the reciprocal of the Period.

$$f = \frac{1}{T} = \frac{\omega}{2\pi}$$

Units: Hertz (Hz) or \(\text{s}^{-1}\).

2.4 Maximum Velocity and Acceleration

These maximum values occur at specific points in the cycle and are crucial for exam questions.

Maximum Velocity (\(v_{max}\))

The speed is maximum when the particle passes through the centre of oscillation (\(x=0\)).

$$v_{max} = A\omega$$

Maximum Acceleration (\(a_{max}\))

Acceleration is maximum at the extremes of the motion (\(x = \pm A\)), where the restoring force is strongest.

$$a_{max} = A\omega^2$$

The Speed-Displacement Equation (V-X Relation)

We often need to find the speed (\(v\)) at any given displacement (\(x\)) without knowing the time (\(t\)). This equation is derived using the identity \(\sin^2\theta + \cos^2\theta = 1\):

$$v^2 = \omega^2 (A^2 - x^2)$$

Memory Aid: This equation is a rearrangement of the Maximum Velocity definition. Notice that if \(x=0\), then \(v^2 = \omega^2 A^2\), so \(v = A\omega\).

Key Takeaway: SHM is defined by \(a = -\omega^2 x\). This simple equation determines the entire motion, including the period \(T\) and the maximum speed \(A\omega\).

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Section 3: Applying SHM to Forces (The Mechanics Link)

In M3, we rarely just solve given differential equations. We usually start with a physical system (like a mass on a spring or an elastic string) and have to prove that the motion is SHM, and then calculate \(\omega\).

3.1 The Standard Derivation Process

To prove that a system exhibits SHM, you must use Newton's Second Law (\(F=ma\)) and show that the resulting equation can be rearranged into the form \(a = -(\text{constant})x\).

Step 1: Identify the Fixed Point \(O\)

Find the point where the net force on the particle is zero. This is the centre of oscillation.

Step 2: Set up the Coordinate System

Let \(x\) be the displacement from the fixed point \(O\).

Step 3: Apply \(F=ma\)

Resolve the forces acting on the particle in the direction of motion. Remember: force towards \(O\) is negative, force away from \(O\) is positive (if \(x\) is positive in that direction).

Step 4: Isolate \(a\)

Rearrange the equation into the form \(a = -(\text{some expression})x\).

Step 5: Identify \(\omega^2\)

The expression in the parentheses is \(\omega^2\). (Since \(\omega\) must be a real number, this expression must be positive.)

3.2 Example: Mass Attached to an Elastic Spring

Consider a mass \(m\) attached to a horizontal spring with stiffness (modulus of elasticity) \(\lambda\) and natural length \(L\). The mass is pulled and released.

1. Restoring Force: The spring force is governed by Hooke's Law: \(F = \frac{\lambda x}{\text{natural length}}\).
2. System Definition: Since we are usually dealing with a simple spring or elastic string where the extension is equal to the displacement \(x\) from the equilibrium point, the restoring force is often just proportional to \(x\) (e.g., \(F = -kx\)).
3. Applying \(F=ma\): If \(F_{\text{net}} = -kx\), then \(ma = -kx\).
4. Isolating \(a\): \(a = -\left(\frac{k}{m}\right) x\).
5. Identifying \(\omega^2\): The motion is SHM with \(\omega^2 = \frac{k}{m}\).

The value of \(k\) might be a complex expression involving \(\lambda, L,\) and \(m\), but the principle remains the same.

3.3 Common Mistakes and Traps

1. Forgetting the Minus Sign: \(a = \omega^2 x\) is *not* SHM. That means the particle accelerates away from the centre, leading to explosion, not oscillation. Always ensure your derivation yields \(a = -\omega^2 x\).
2. Misidentifying the Centre \(O\): If the problem involves gravity (e.g., a vertical spring), the centre of oscillation \(O\) is NOT the natural length position. It is the equilibrium position where the tension/thrust balances the weight \(mg\). All displacements \(x\) must be measured from this equilibrium point.
3. Mixing up \(A\) and \(x\): \(A\) (Amplitude) is the maximum distance travelled from the centre \(O\). \(x\) is the displacement at any given time. Often, the amplitude \(A\) must be calculated using the release condition (e.g., if released from rest at \(x=0.5\text{m}\), then \(A=0.5\text{m}\)).

Quick Tip for Vertical Springs (Gravitational Systems)
When solving vertical problems, draw a clear diagram showing the natural length, the equilibrium point (extension \(e\)), and the general displacement \(x\) (measured from the equilibrium point). The beauty of SHM is that once you find the equilibrium, the \(mg\) and initial tension forces cancel out, and you are usually left with a simple form of Hooke's Law proportional to \(x\)!

Key Takeaway: The process of proving SHM is mechanical: use \(F=ma\), find the centre \(O\), and algebraically manipulate the resulting equation into the standard form \(a = -\omega^2 x\).

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Section 4: Energy in SHM (Connection to M3 Work, Energy & Power)

Kinematics often connects directly to the M3 chapter on Work, Energy, and Power. The total mechanical energy in an SHM system (assuming no damping) is always conserved.

4.1 Energy Components in SHM

The system energy shifts between two main forms:

1. Kinetic Energy (KE): Energy of motion, \(KE = \frac{1}{2} mv^2\). This is maximum at the centre (\(x=0\)) and zero at the extremes (\(x = \pm A\)).
2. Potential Energy (PE): Energy stored in the system (e.g., Elastic Potential Energy, EPE). This is maximum at the extremes and zero at the centre.

Total Energy (E) = KE + PE = Constant

4.2 Using the V-X Relation for Energy

If we substitute the speed-displacement equation \(v^2 = \omega^2 (A^2 - x^2)\) into the KE equation:

$$KE = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)$$

If the particle is at an extreme (\(x=A\)), KE = 0, meaning all energy is PE.

If the particle is at the centre (\(x=0\)), PE = 0, and KE is maximum:

$$KE_{max} = \frac{1}{2} m \omega^2 A^2$$

Since the total energy is conserved and equal to \(KE_{max}\), we can deduce the total energy of the system:

$$E_{total} = \frac{1}{2} m \omega^2 A^2$$

This formula is immensely useful for finding the amplitude \(A\) if you know the maximum speed, or vice versa.

Key Takeaway: Maximum kinetic energy in SHM is \(KE_{max} = \frac{1}{2} m \omega^2 A^2\). This represents the total mechanical energy of the system.

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Conclusion and Final Encouragement

Further Kinematics might seem abstract due to the high use of symbols (\(\omega\), \(A\), \(\lambda\)), but remember that every variable has a physical meaning. Simple Harmonic Motion is a direct consequence of a linear restoring force, and the core of the chapter is proving \(a = -\omega^2 x\) for various physical setups.

Practice separating variables when using \(a = v \frac{dv}{dx}\), and always draw a clear diagram to identify your origin \(O\) before applying \(F=ma\). You've got this!

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