Oscillations

16 Thermodynamics – Internal Energy and Work

16.1 Internal Energy (U)

• Definition: the total microscopic kinetic + potential energy of all particles in a system.
• State function: U depends only on the current state (T, p, V, n…) and not on the path taken to reach that state.
• Kinetic‑theory view (ideal gas): for a monatomic ideal gas the microscopic kinetic energy of the molecules gives

  \$U = \tfrac{3}{2}\,nRT\$

  where n is the amount of gas, R the universal gas constant and T the absolute temperature.

• Temperature dependence: for any ideal gas

  \$U = nC_{V}T\$

  where C_V is the molar heat capacity at constant volume.

  Hence a change in temperature directly changes the internal energy:

  \$\Delta U = nC_{V}\,\Delta T\$

• Heat‑capacity relationship (Cambridge 9702):

  \$C{P}=C{V}+R\$

  (useful when a process occurs at constant pressure).

• Independence of pressure/volume (ideal gas): because U is a function of T only, changing p or V at constant T does not change U.

16.2 Work Done at Constant Pressure

When a gas expands or contracts against a constant external pressure p, the work done on the system is

\$W = -p\Delta V\$

• Cambridge sign convention: W is positive when work is done on the system (compression) and negative when the system does work on the surroundings (expansion).
• First‑law form used in the syllabus:

  \$\Delta U = q + W\$

  where q is the heat added to the system (positive when heat flows into the system).

First Law of Thermodynamics (syllabus form)

\$\boxed{\Delta U = q + W}\$

• Isochoric process (ΔV = 0): \(W=0\) ⇒ \(\Delta U = q\).
• Isobaric process (constant p): substitute \(W = -p\Delta V\) into the first‑law equation.
• Isothermal ideal‑gas expansion: \(\Delta U = 0\) (since \(T\) constant) ⇒ \(q = -W = p\Delta V\).

17 Oscillations – Simple Harmonic Motion (SHM) and Damping

17.1 Simple Harmonic Oscillator (SHO) – Core Equations

Energy in an Undamped SHO

• Kinetic energy

  \$K = \tfrac12 mv^{2}= \tfrac12 m\omega^{2}\bigl(x_{0}^{2}-x^{2}\bigr)\$

• Elastic potential energy of the spring

  \$U_{s}= \tfrac12 kx^{2}= \tfrac12 m\omega^{2}x^{2}\$

• Total mechanical energy (constant)

  \$E{\text{mech}} = K+U{s}= \tfrac12 m\omega^{2}x{0}^{2}= \tfrac12 kx{0}^{2}\$

17.2 Damped Oscillations

A real oscillator experiences a resistive force that opposes the motion. In the Cambridge syllabus the damping force is taken as proportional to the velocity:

\$F_{\text{damp}} = -b\,v\$

• b – damping constant (units kg s⁻¹). The negative sign ensures the force always acts opposite to the direction of motion.
• The damping force does negative work on the oscillator, removing mechanical energy and converting it to heat in the surroundings.

Power and Energy Loss

The instantaneous power supplied by the damping force is

\$P = F_{\text{damp}}\,v = -b\,v^{2}\$

• The negative sign shows that mechanical energy is being removed at a rate \(|P|\).
• Applying the first law to the combined system (oscillator + surroundings):

  \$\Delta U{\text{sur}} = q = -W{\text{damp}} = b\,v^{2}\$

  where \(\Delta U_{\text{sur}}\) is the increase in internal energy (heat) of the surroundings.

Types of Damping (Cambridge 9702 17.3)

Mathematical form (optional for A‑Level)

For a linear damped SHO the displacement can be written as

\$\$x(t)=A\,e^{-\,\frac{b}{2m}t}\sin(\omega' t+\phi),\qquad

\omega'=\sqrt{\omega^{2}-\left(\frac{b}{2m}\right)^{2}}\$\$

where \(\omega' \) is the damped angular frequency. When \(b/2m \ll \omega\) the motion is essentially the same as the undamped case.

17.3 Resonance (Driven Oscillations)

• A periodic driving force \(F{\text{drive}} = F{0}\sin(\omega_{d}t)\) is applied to the oscillator.
• The steady‑state amplitude is

  \$A(\omega{d}) = \frac{F{0}}{m\sqrt{(\omega^{2}-\omega{d}^{2})^{2}+(\frac{b\omega{d}}{m})^{2}}}\$

• Resonance condition: the amplitude reaches a maximum when the driving angular frequency \(\omega_{d}\) equals the natural (undamped) angular frequency \(\omega\) (or is very close to it for light damping).
• At resonance the phase difference between the driving force and the displacement is \(\pi/2\) rad (90°).
• In the syllabus this is identified by a peak in an amplitude‑versus‑frequency graph.

17.4 Models of Physical Systems

• Mass‑spring system: restoring force \(F=-kx\).
• Simple pendulum (small angles): restoring torque \(\tau\approx -mgL\,\theta\) ⇒ SHM with \(\omega=\sqrt{g/L}\).
• LC electrical circuit: charge‑displacement analogue; restoring “force” provided by the inductor, \(\omega=1/\sqrt{LC}\).
• Damping models:

  • Air resistance (linear: \(-b v\)) – syllabus approved approximation.
  • Viscous friction in fluids.
  • Electrical resistance in an RLC circuit (analogue of \(-b v\)).

Energy Transfer Summary (Thermodynamics ↔ Oscillations)

Sample A‑Level Question (Cambridge style)

Question: A mass \(m = 0.5\;\text{kg}\) is attached to a spring of constant \(k = 200\;\text{N m}^{-1}\). The system oscillates in air with a damping force \(F_{\text{damp}} = -0.05\,v\). The amplitude at \(t = 0\) is \(A = 0.10\;\text{m}\). Calculate the rate at which mechanical energy is being converted to heat when the mass passes through the equilibrium position for the first time.

Solution (step‑by‑step, AO2 style)

1. Angular frequency (ignoring damping for the speed at equilibrium):

   \[

   \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = 20\;\text{rad s}^{-1}

   \]

2. Maximum speed at the equilibrium position:

   \[

   v_{\max}= \omega A = 20 \times 0.10 = 2.0\;\text{m s}^{-1}

   \]

3. Power dissipated by the damping force:

   \[

   P = -F_{\text{damp}}\,v = -(-0.05\,v)\,v = 0.05\,v^{2}

   \]

4. Insert \(v = v_{\max}\):

   \[

   P = 0.05 \times (2.0)^{2} = 0.20\;\text{W}

   \]

5. Interpretation: 0.20 J of mechanical energy is converted to heat each second when the mass passes through the equilibrium point.

Practical Skills Linked to the Syllabus

• Measuring period and frequency: Use a stopwatch, photogate, or motion‑sensor software to record displacement‑time data and obtain \(T\) and \(f\).
• Determining the damping constant \(b\): Record successive amplitudes \(A{1},A{2},\dots\); plot \(\ln A\) against time. The gradient equals \(-b/(2m)\) (linear‑fit method).
• Heat measurement in a damped system: Place a small calorimeter or thermistor in the surrounding air; the temperature rise \(\Delta T\) gives the heat \(q = C_{\text{cal}}\Delta T\), confirming the energy‑loss calculation.
• Resonance experiment: Drive a mass‑spring system with a function generator; vary the driving frequency and plot amplitude versus frequency to locate the resonance peak.

Assessment Objectives Covered

• AO1 – Knowledge: Definitions of internal energy, heat capacities, SHM equations, damping types, resonance.
• AO2 – Application: Use \(\Delta U = q + W\), \(F_{\text{damp}}=-b v\), and energy‑conservation formulas to solve quantitative problems.
• AO3 – Analysis: Interpret displacement‑time graphs to distinguish light, critical and heavy damping; analyse amplitude‑frequency data for resonance.

Key Points to Remember

• Internal energy is the microscopic kinetic + potential energy; for an ideal gas it depends only on temperature (U ∝ T).
• Heat capacities: \(C{P}=C{V}+R\); use \(C{V}\) for constant‑volume processes and \(C{P}\) for constant‑pressure processes.
• Cambridge sign convention in the first law: \(\Delta U = q + W\) (both \(q\) and \(W\) positive when they add energy to the system).
• In a SHO without non‑conservative forces, total mechanical energy is constant: \(E{\text{mech}}=\tfrac12 kx{0}^{2}\).
• Damping removes mechanical energy at a rate \(P=-b v^{2}\); the lost energy appears as heat, raising the internal energy of the surroundings.
• Resonance occurs when the driving frequency equals the natural frequency; the amplitude peaks and the phase difference is \(\pi/2\) rad.