Welcome to Further Dynamics! Mastering Variable Forces and Oscillations
Welcome to the M3 section on Further Dynamics! If you felt comfortable with Newton’s Laws in M1 and M2, get ready for the next level. In M3, forces are no longer constant—they change depending on where the object is, how fast it is going, or what the time is.
This chapter is all about combining the power of calculus (integration and differentiation) with mechanics to solve complex, realistic motion problems, especially those involving elasticity and oscillations.
Don't worry if this seems tricky at first. By breaking down the complex relationships and choosing the right form of acceleration, we can tackle any problem!
Section 1: Dynamics of Variable Force
The foundation of all dynamics is still Newton's Second Law: \(F = ma\). However, when the force \(F\) is variable, we must express the acceleration \(a\) using calculus, which allows us to relate force, velocity, displacement, and time.
1.1 Choosing the Right Form of Acceleration
The key to solving variable force problems is recognizing what the force \(F\) is a function of (t, x, or v) and then selecting the appropriate expression for acceleration.
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If \(F\) depends on Time (\(t\)):
We use \(a = \frac{dv}{dt}\). This is a differential equation where we can separate the variables to find velocity or time.
\[ F(t) = m \frac{dv}{dt} \implies \int F(t) \, dt = \int m \, dv \] -
If \(F\) depends on Displacement (\(x\)):
We use the chain rule form \(a = v \frac{dv}{dx}\). This is often used when dealing with work and energy, as integrating \(F\) with respect to \(x\) gives us work done.
\[ F(x) = m v \frac{dv}{dx} \implies \int F(x) \, dx = \int m v \, dv \]Note: The term on the right, \(\int m v \, dv\), integrates to \(\frac{1}{2} m v^2\), which is Kinetic Energy (KE). This confirms that the work done by the force changes the KE.
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If \(F\) depends on Velocity (\(v\)):
This often occurs when resistance or drag forces are involved (e.g., air resistance proportional to \(v\) or \(v^2\)). You might need to find velocity as a function of time or displacement.
To find velocity as a function of time: \[ F(v) = m \frac{dv}{dt} \implies \int dt = \int \frac{m}{F(v)} \, dv \] To find velocity as a function of displacement: \[ F(v) = m v \frac{dv}{dx} \implies \int dx = \int \frac{m v}{F(v)} \, dv \]
Quick Review: Key Calculus Forms for \(a\)
\(a\) is a function of \(x\): \(v \frac{dv}{dx}\) (Finds \(v(x)\))
\(a\) is a function of \(t\) or \(x\): \(\frac{d^2x}{dt^2}\)
1.2 Step-by-Step Problem Solving (Variable Force)
- Identify: Determine what variable (\(t, x\), or \(v\)) the force \(F\) depends on.
- Choose: Select the appropriate form of \(a\) (\(\frac{dv}{dt}\) or \(v \frac{dv}{dx}\)).
- Set up: Substitute into \(F = ma\).
- Separate: Rearrange the equation so all terms involving one variable are on one side (e.g., all \(v\) terms with \(dv\), and all \(t\) terms with \(dt\)).
- Integrate: Apply definite or indefinite integration, using the initial conditions (boundary conditions) to find the constant of integration if needed.
Key Takeaway: Variable force problems rely entirely on choosing the correct calculus relationship for acceleration and executing the integration accurately.
Section 2: Elastic Strings and Springs
This section introduces two new concepts that store potential energy: Elastic Strings and Springs. The mathematics used to describe both is identical.
2.1 Hooke’s Law: The Tension Formula
Hooke’s Law states that the tension (or thrust, for compression) in an elastic string or spring is proportional to its extension (or compression), provided the elastic limit is not exceeded.
The key formula for Tension \(T\) is: \[ T = \frac{\lambda x}{L} \]
- \(\lambda\) (Lambda): The Modulus of Elasticity. This is a measure of stiffness. A large \(\lambda\) means a stiff spring/string. Units are typically Newtons (N).
- \(x\): The extension or compression (difference between the current length and the natural length). Always positive.
- \(L\): The natural length of the string or spring when no force is applied.
Did you know? (Strings vs. Springs)
Elastic Strings can only exert tension when they are stretched (\(x > 0\)). They go slack (T = 0) if the length is less than the natural length. Elastic Springs, however, can be stretched (Tension) or compressed (Thrust/Compression Force). In M3 calculations, we often treat the thrust in a compressed spring as 'negative tension'.
2.2 Elastic Potential Energy (EPE)
When you stretch a spring, you are doing work against the tension force. This stored energy is the Elastic Potential Energy (EPE).
Since EPE is the work done stretching the spring/string (Work \(W = \int T \, dx\)), we integrate Hooke’s Law: \[ EPE = \int_0^x T \, dx = \int_0^x \frac{\lambda s}{L} \, ds \]
The resulting formula for EPE is: \[ EPE = \frac{\lambda x^2}{2L} \]
Crucial Point: EPE is calculated using the extension \(x\). Always find \(x\) first!
2.3 Conservation of Energy (The Complete Picture)
When an object attached to an elastic string moves (especially vertically), we must include EPE in our conservation of energy equation alongside Kinetic Energy (KE) and Gravitational Potential Energy (GPE).
The Full Energy Equation (M3): \[ KE_{initial} + GPE_{initial} + EPE_{initial} = KE_{final} + GPE_{final} + EPE_{final} \]
- KE: \(\frac{1}{2} m v^2\)
- GPE: \(m g h\) (Remember to define a zero potential energy level!)
- EPE: \(\frac{\lambda x^2}{2L}\)
Memory Aid: Think of energy storage containers. KE is movement, GPE is height, and EPE is stretchiness. When one goes down, others must go up.
Key Takeaway: Hooke's Law defines tension (\(T = \frac{\lambda x}{L}\)) and the integration of this law defines the stored energy (\(EPE = \frac{\lambda x^2}{2L}\)). Use the full conservation of energy equation to solve motion problems.
Section 3: Simple Harmonic Motion (SHM)
Simple Harmonic Motion describes the smooth, repetitive oscillation observed in many physical systems, such as a mass on a vertical spring, or a simple pendulum (for small angles).
3.1 Defining Characteristics of SHM
An object undergoes SHM if its acceleration is always proportional to its displacement from a fixed equilibrium point, and is always directed towards that point.
The defining differential equation is: \[ \frac{d^2x}{dt^2} = - \omega^2 x \]
- \(x\): Displacement from the equilibrium position (not the fixed point).
- \(\omega^2\) (Omega squared): A positive constant, often derived from physical parameters like mass, stiffness, and gravity. \(\omega\) is the angular frequency (measured in rad s-1).
- The negative sign (\( - \)): This is essential. It means if \(x\) is positive (to the right), the acceleration is negative (to the left), pulling it back to the origin.
3.2 Standard Equations of SHM
The general solution to the SHM differential equation is based on sine and cosine functions.
If the motion starts at the equilibrium position (\(x=0\) at \(t=0\)): \[ x = a \sin(\omega t) \]
If the motion starts at maximum displacement (at the amplitude \(x=a\) at \(t=0\)): \[ x = a \cos(\omega t) \]
where \(a\) is the Amplitude (the maximum displacement from the equilibrium position).
Key Formulas derived from \(x\)
By differentiating the position equation \(x(t)\), we get velocity \(v(t)\) and acceleration \(a(t)\).
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Velocity \(v\) (Maximum Speed)
The velocity at any point \(x\) is given by: \[ v^2 = \omega^2 (a^2 - x^2) \] The Maximum Speed occurs when \(x=0\) (at the equilibrium point): \[ v_{max} = a\omega \]
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Acceleration \(a\) (Maximum Acceleration)
The maximum magnitude of acceleration occurs when \(x=\pm a\): \[ a_{max} = \omega^2 a \]
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Period and Frequency
The Period \(T\) (time taken for one full oscillation) and Frequency \(f\) (oscillations per second) are determined by \(\omega\): \[ T = \frac{2\pi}{\omega} \hspace{1cm} \text{and} \hspace{1cm} f = \frac{1}{T} = \frac{\omega}{2\pi} \]
3.3 How to Show a System Exhibits SHM
To prove a system exhibits SHM, you must demonstrate that the net restoring force \(F_{net}\) results in the form \(a = -\omega^2 x\).
Step-by-Step Derivation for a Mass on a Vertical Spring
Consider a mass \(m\) suspended from a spring (natural length \(L\), modulus \(\lambda\)).
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Find the Equilibrium Position (\(E\)):
At equilibrium, acceleration is zero. The forces balance: Tension \(T_0 = mg\).
Using Hooke's Law: \(mg = \frac{\lambda e}{L}\), where \(e\) is the extension at equilibrium. -
Consider Motion away from Equilibrium:
Let the mass be pulled down a further displacement \(x\) from \(E\).
Total extension is now \(e+x\).
New Tension \(T_{new} = \frac{\lambda (e+x)}{L}\). -
Apply \(F=ma\):
We take the downward direction as positive. The net force is (Down Force - Up Force): \[ F_{net} = mg - T_{new} \] \[ m \frac{d^2x}{dt^2} = mg - \frac{\lambda (e+x)}{L} \] \[ m \frac{d^2x}{dt^2} = mg - \frac{\lambda e}{L} - \frac{\lambda x}{L} \]
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Simplify using the Equilibrium Condition:
From Step 1, we know \(mg = \frac{\lambda e}{L}\). These terms cancel out! \[ m \frac{d^2x}{dt^2} = 0 - \frac{\lambda x}{L} \] \[ \frac{d^2x}{dt^2} = - \left( \frac{\lambda}{mL} \right) x \]
This is in the form \(\frac{d^2x}{dt^2} = - \omega^2 x\).
Therefore, the motion is SHM, and the angular frequency is:
\[ \omega = \sqrt{\frac{\lambda}{mL}} \]
Common Mistake to Avoid
When solving SHM problems involving springs, always measure displacement \(x\) from the equilibrium position, not from the natural length point. If you measure from the wrong point, the \(mg\) and \(\frac{\lambda e}{L}\) terms will not cancel out, and you won't get the standard SHM equation.
Key Takeaway: SHM is defined by \(a = -\omega^2 x\). Once you find \(\omega^2\), all key parameters (Period, Max Speed, Max Acceleration) are immediately calculable.
Summary of Further Dynamics (M3)
Further Dynamics combines the physical principles you know with the advanced tools of calculus.
Variable Forces: Choose \(a = \frac{dv}{dt}\) (for F(t) or F(v)) or \(a = v \frac{dv}{dx}\) (for F(x) or F(v)). Integration is essential here.
Elasticity: Remember the two key formulae for elastic strings/springs:
1. Hooke's Law (Force): \(T = \frac{\lambda x}{L}\)
2. Elastic Potential Energy (Energy): \(EPE = \frac{\lambda x^2}{2L}\)
SHM: Look for the defining equation \(\frac{d^2x}{dt^2} = - \omega^2 x\). If you can manipulate the net force equation \(F=ma\) into this form, you have found \(\omega\).
Keep practicing those integrations and remember your definitions! You’ve got this!