Resonance and applications of waves

Review harmonic motion and natural frequency

An ideal spring exerts restoring force F=−kx, where x is displacement from equilibrium and k is stiffness in N/m. The negative sign indicates the direction opposite to displacement. By the second law, mx″=−kx. The function x=A cos(ω₀t+φ) satisfies this equation if ω₀²=k/m. Natural angular frequency is ω₀=√(k/m), in rad/s, and frequency in hertz is f₀=ω₀/(2π).

Amplitude A is the maximum displacement from equilibrium; phase φ indicates the initial state. Two oscillations can have the same frequency and different amplitudes, or the same amplitude and different phases. Ideal mechanical energy alternates between kinetic and elastic potential energy, with total kA²/2. The model is linear: it assumes a spring in its proportional elastic regime and constant effective mass. Large displacements or materials outside that regime may require another model.

Damping: energy leaves the motion

A viscous damping force can be approximated by −bx′, with b in kg/s. Without external excitation, the equation is mx″+bx′+kx=0. This force's power is −b(x′)²≤0, showing that it removes mechanical energy and transfers it into other forms. This explains decreasing amplitude without claiming that total energy disappears.

With weak damping, the system continues oscillating and amplitude decays. In the underdamped case, damped free angular frequency is √(ω₀²−(b/2m)²), lower than ω₀. If b=2√(mk), damping is critical, the boundary between oscillation and nonoscillatory return; above that, return is overdamped. Behavior also depends on initial conditions. Undamped natural frequency and damped free frequency should not be confused just because they can be close when b is small.

A periodic force changes the question

Now apply F(t)=F₀ cos(ωt). The equation becomes mx″+bx′+kx=F₀ cos(ωt). Frequency ω is chosen by the external agent, whereas ω₀ belongs to the mass–spring model. The initial response includes a transient depending on initial conditions. With positive damping, this transient decays, leaving a steady state at the driving frequency.

In steady state, we write x=A(ω)cos(ωt−δ), where δ is phase lag. At low frequencies, the spring follows the force almost quasistatically. At very high frequencies, inertia inhibits large displacements and the response lags by almost half a cycle. Large amplitude may occur between these regimes. Growth is not creation of energy: the external force supplies work each cycle and damping dissipates energy. Steady state is established when the averages of these transfers balance.

Where amplitude comes from and where the peak lies

Substituting the sinusoidal response into the equation produces one part proportional to displacement and another in quadrature, associated with velocity. Their coefficients are (k−mω²)A and bωA. Since these parts differ in phase by a quarter-cycle, force amplitude satisfies F₀²=A²[(k−mω²)²+(bω)²]. Thus:

The largest displacement depends on the entire denominator, not only k−mω². For fixed force amplitude and viscous damping, the displacement peak occurs at ωp=√(ω₀²−b²/(2m²)), when b²<2mk. If damping is greater, there is no such positive interior peak. Therefore no universal rule places every amplitude maximum exactly at ω₀. With weak damping, ωp lies close to ω₀ and that approximation is often suitable.

Phase and energy: which response are we measuring?

Phase satisfies tanδ=bω/(k−mω²), choosing the correct quadrant for 0≤δ≤π. At ω=ω₀, δ=π/2 for b>0: displacement lags force by a quarter-cycle. Velocity is in phase with force and average energy transfer is efficient. Using only the arctangent without considering the denominator's sign can produce the wrong phase above ω₀.

Average dissipated power is bω²A²/2. For this model and fixed F₀, it reaches its maximum at ω=ω₀ even when the displacement peak lies slightly below. This difference teaches us to state which quantity is observed: displacement, velocity or power. Greater damping reduces and broadens the displacement response. A narrow peak selects frequencies but can make the system sensitive to small changes. Without damping and with excitation exactly at the natural frequency, there is no finite steady amplitude: the ideal model predicts growth that real systems eventually limit.

Standing waves and modes

On a string with fixed ends, reflection and superposition allow modes with nodes at the ends. n half-wavelengths fit along L: λₙ=2L/n, with n a positive integer. If wave speed is v, fₙ=nv/(2L). The fundamental mode has n=1; the following modes are integer harmonics in this ideal uniform-string model.

Excitation near a mode can produce a strong response if it couples to that pattern and dissipation does not suppress it. Matching a frequency alone is not enough: the force application position and symmetry matter. Driving at a node of a particular mode may couple weakly to it. Air columns with different end conditions have different mode sequences. Musical instruments and vibration-control systems exploit or limit these responses, always respecting geometry, dissipation and operating amplitude.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

Resonance makes energy appear from nothing.

Excitation supplies energy; dissipation and nonlinearity limit the response.

Natural frequency and every amplitude peak are identical.

The peak depends on damping, excitation and the quantity observed.

A standing wave means no motion anywhere on the string.

Nodes remain still, but antinodes oscillate; the spatial pattern remains.