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Energy and Resonance
Energy Conservation in SHM
When a mass $m$ is displaced by $x_o$ against the restoring force of a spring, work is done that stores as elastic potential energy. This derivation uses the average force method.
$$W = \bar{F} \cdot d = \frac{1}{2}kx_o \times x_o = \frac{1}{2}kx_o^2$$
Work done against the spring equals the elastic potential energy stored.
$\bar{F}$=Average force during displacement(N)
$k$=Spring constant(N/m)
$x_o$=Maximum displacement (amplitude)(m)
Slow Stretching: The derivation assumes the spring is stretched slowly (zero acceleration), so the applied force equals the restoring force at every point.
Linear Force: Since Hooke's law gives $F = kx$, the force-displacement graph is a straight line — the work done is the triangular area under it.
Energy in Simple Harmonic Motion continuously oscillates between kinetic energy and elastic potential energy while the total mechanical energy remains constant.
$$E_{total} = \frac{1}{2}kA^2 = \text{constant}$$
The law of conservation of energy applied to an oscillating system.
$E_{total}$=Total mechanical energy(J)
$k$=Spring constant(N/m)
$A$=Amplitude of oscillation(m)
At equilibrium ($x=0$)
→Total energy is entirely kinetic: $E_{total} = \frac{1}{2}mv_{max}^2$
At amplitude ($x=\pm A$)
→Total energy is entirely potential: $E_{total} = \frac{1}{2}kA^2$
Equilibrium Position: Velocity is maximum and restoring force is zero, making K.E. maximum and P.E. zero.
Maximum Displacement: Velocity is zero and restoring force is maximum, making K.E. zero and P.E. maximum.
Pendulum Analogy: In a simple pendulum, gravitational P.E. at the top of the swing converts to K.E. at the bottom — same interchange, different energy type.
Proportionality Shortcut: Since $E \propto A^2$, if amplitude drops by half, energy drops to one-quarter.
At any displacement $x$, the energy splits into kinetic energy (motion) and elastic potential energy (deformation), and their sum always equals the total.
$$P.E. = \frac{1}{2}kx^2 \quad ; \quad K.E. = \frac{1}{2}k(A^2 - x^2)$$
The position-dependent energy partition in SHM.
$x$=Instantaneous displacement from equilibrium(m)
$A$=Amplitude(m)
$k$=Spring constant(N/m)
$x = A/2$
→P.E. = $\frac{1}{4}E_{total}$, K.E. = $\frac{3}{4}E_{total}$
$x = A/\sqrt{2}$
→P.E. = K.E. = $\frac{1}{2}E_{total}$ (the equal-split point)
Ratio Shortcut: To find P.E. at any position, just compute $(x/A)^2$ and multiply by $E_{total}$. No need to know $k$ or $m$ separately.
Velocity at Position x: From K.E. = $\frac{1}{2}mv^2$, the velocity is $v = v_{max}\sqrt{1 - x^2/A^2}$.
Dimensional Check: Both $\frac{1}{2}kx^2$ and $\frac{1}{2}mv^2$ have dimensions $[M][L]^2[T]^{-2}$ = Joules.
Energy-Displacement Graphs
The relationship between energy and displacement follows a parabolic relationship, with the total energy represented by a horizontal line.
$$K.E. = \frac{1}{2}k(A^2 - x^2)$$
Kinetic energy as a function of position — an inverted parabola.
$k$=Spring constant(N/m)
$A$=Amplitude(m)
$x$=Instantaneous displacement(m)
$x = 0$ (equilibrium)
→K.E. = $\frac{1}{2}kA^2$ = maximum
$x = \pm A$ (extremes)
→K.E. = 0, object momentarily at rest
P.E. Curve: Upward-opening parabola ($\frac{1}{2}kx^2$) — minimum at center, maximum at edges.
K.E. Curve: Downward-opening parabola — maximum at center, zero at edges.
Equal Energy Point: At $x = \pm A/\sqrt{2} \approx \pm 0.707A$, the K.E. and P.E. curves cross. This is the only position where energy is split equally.
Total Energy Line: A flat horizontal line at $E = \frac{1}{2}kA^2$, confirming conservation.
Free and Forced Oscillations
A body executing free oscillations vibrates without interference from an external force, oscillating at its natural frequency determined solely by the system's physical properties.
Natural Frequency: For a spring-mass system, $f_0 = \frac{1}{2\pi}\sqrt{k/m}$. For a pendulum, $f_0 = \frac{1}{2\pi}\sqrt{g/l}$.
Independence from Amplitude: The natural frequency of ideal SHM does not depend on amplitude — only on system parameters ($k$, $m$, or $l$, $g$).
Real Systems: Free oscillations gradually die out due to friction and air resistance dissipating energy.
When a freely oscillating system is subjected to an external periodic force, forced oscillations take place. The system is then called a driven harmonic oscillator.
Driving Force: An external periodic force that continuously supplies energy to the system.
Steady State: After initial transients die out, the system oscillates at the driving frequency, not its natural frequency.
Examples: Vehicle vibrations from engine, loud music from wooden boards of string instruments, pushing a swing periodically.
Resonance
Resonance occurs when the driving frequency of an external periodic force equals the natural frequency of the oscillator, causing the amplitude to become extraordinarily large.
$$f_{driving} = f_0$$
The condition for maximum energy transfer and maximum amplitude.
$f_{driving}$=Frequency of the external periodic force(Hz)
$f_0$=Natural frequency of the system(Hz)
$f_{driving} \ll f_0$
→Amplitude is small — the system barely responds to the slow driving
$f_{driving} = f_0$
→Maximum amplitude — energy transfer from driving source is most efficient
$f_{driving} \gg f_0$
→Amplitude is small — the system cannot keep up with the fast driving
Maximum Energy Transfer: At resonance, the driving force is perfectly synchronized with the oscillation — each push adds energy at the optimal moment.
Phase at Resonance: The velocity of the oscillator is in phase with the driving force, maximizing power transfer.
Coupled Pendulum Demo: Pendulums of matching length on a shared rod transfer energy efficiently; non-matching ones remain small — a classic demonstration of resonance selectivity.
Resonance has both beneficial applications and dangerous consequences in real-world systems.
Radio Tuning: Turning the dial changes the circuit's natural frequency to match the broadcast frequency — at resonance, energy absorption from the signal is maximum and only that station is heard.
Microwave Oven: Produces waves at 2450 MHz (wavelength 12 cm) that match the resonant frequency of water and fat molecules, causing them to absorb energy and heat the food.
Soldiers on Bridge: Marching soldiers break step on long bridges because rhythmic marching could match the bridge's natural frequency and build up dangerously large oscillations.
Tacoma Narrows Bridge: Collapsed due to wind-driven oscillations matching the bridge's natural frequency — a dramatic example of destructive resonance.
Damped Oscillations
Damped oscillations are oscillations whose amplitude decreases steadily with time because resistive forces (friction, air resistance, viscous drag) dissipate energy from the system.
$$E \propto A^2$$
Since energy is proportional to amplitude squared, any reduction in amplitude means energy has been lost to resistive forces.
$E$=Remaining mechanical energy(J)
$A$=Current amplitude (decreasing over time)(m)
Ideal vs Real: Pure SHM (constant amplitude forever) is an idealization. All real oscillators experience damping.
Light Damping: System oscillates many times with a slowly decaying exponential decay amplitude envelope.
Critical Damping: System returns to equilibrium in the shortest possible time without overshooting — used in car shock absorbers to quickly suppress oscillations after a bump.
Overdamping: Very high resistance causes an extremely slow, non-oscillatory return to equilibrium.
Limiting Case: As damping → 0, the system approaches ideal SHM with constant amplitude.
Sharpness of Resonance
The sharpness of resonance describes how rapidly the amplitude drops when the driving frequency deviates from the natural frequency. It depends inversely on damping.
Sharp Resonance (low damping): Tall, narrow peak — amplitude is very large at $f_0$ but drops off rapidly. A lead-bob pendulum shows this behavior due to low air resistance relative to its weight.
Flat Resonance (high damping): Short, wide peak — moderate amplitude over a broad frequency range. A pith-ball pendulum shows this due to high air resistance relative to its light weight.
Practical Implication: In radio circuits, sharp resonance allows tuning to a single station without interference from neighboring frequencies.
Application Strategy
Worked Examples
Problem Statement
A 0.5 kg mass is attached to a spring with k = 200 N/m. It is pulled 10 cm from equilibrium and released. Calculate the maximum speed of the mass.
Conservation of EnergySHM Energy
Strategic Toolkit
$$E_{total} = \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$$
Step 1
Identify Variables
Convert all units to SI: m = 0.5 kg, k = 200 N/m, A = 0.1 m.
$A = 10\text{ cm} = 0.1\text{ m}$
Step 2
Calculate Total Energy
All energy is initially Potential Energy at the release point.
$E = \frac{1}{2}(200)(0.1)^2 = 1.0\text{ J}$
Step 3
Solve for Maximum Speed
At equilibrium, all energy is Kinetic. Equate total energy to $\frac{1}{2}mv^2$.
$1.0 = \frac{1}{2}(0.5)v_{max}^2 \implies v_{max} = \sqrt{4} = 2.0\text{ m/s}$
Problem Statement
A spring with $k = 80.0$ N/m vertically supports a 1.0 kg mass. Find the distance by which the mass must be pulled down so that on release it passes the mean position with a velocity of 1.0 m/s.
Angular FrequencyAmplitude-Velocity Relation
Strategic Toolkit
$$v_{max} = A\omega \quad ; \quad \omega = \sqrt{k/m}$$
Step 1
Calculate Angular Frequency
Use the spring constant and mass to find $\omega$.
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{80}{1.0}} = 8.94\text{ rad/s}$
Step 2
Apply Velocity-Amplitude Relation
At the mean position, all energy is kinetic and $v = v_{max} = A\omega$.
$A = \frac{v_{max}}{\omega} = \frac{1.0}{8.94} = 0.11\text{ m}$
Problem Statement
An 8.0 kg body executes SHM with amplitude 30 cm. The restoring force at 30 cm is 60 N. Find the acceleration, speed, K.E. and P.E. when displacement is 12 cm.
Energy at Position xAcceleration in SHMSpring Constant
Strategic Toolkit
$$k = F/x \quad ; \quad a = \omega^2 x \quad ; \quad v = \omega\sqrt{A^2 - x^2}$$
Step 1
Find Spring Constant
From Hooke's law at maximum displacement.
$k = \frac{F}{x} = \frac{60}{0.30} = 200\text{ N/m}$
Step 2
Find Angular Frequency
Use $\omega = \sqrt{k/m}$.
$\omega = \sqrt{\frac{200}{8.0}} = 5.0\text{ rad/s}$
Step 3
Acceleration at x = 12 cm
Acceleration in SHM: $a = \omega^2 x$.
$a = (5.0)^2 \times 0.12 = 3.0\text{ m/s}^2$
Step 4
Speed at x = 12 cm
Use the velocity-displacement relation.
$v = \omega\sqrt{A^2 - x^2} = 5.0\sqrt{0.09 - 0.0144} = 5.0 \times 0.275 = 1.37\text{ m/s}$
Step 5
Energy Values
Calculate K.E. and P.E. at this position.
$K.E. = \frac{1}{2}(8.0)(1.37)^2 = 7.6\text{ J}$ ; $P.E. = \frac{1}{2}(200)(0.12)^2 = 1.44\text{ J}$
Problem Statement
A child on a swing (natural frequency 0.5 Hz) is being pushed. Describe the effect on amplitude if the pushes occur every 1.5 seconds versus every 2.0 seconds.
ResonanceFrequency-Period Relationship
Strategic Toolkit
$$f = 1/T$$
Step 1
Determine Natural Period
Find the period that matches the swing's natural frequency.
$T_0 = \frac{1}{f_0} = \frac{1}{0.5} = 2.0\text{ s}$
Step 2
Analyze Case 1 (1.5s)
The driving period (1.5s) does not match the natural period (2.0s). Energy transfer is inefficient.
$f_{drive} = 1/1.5 = 0.67\text{ Hz} \neq f_0 = 0.5\text{ Hz}$
Step 3
Analyze Case 2 (2.0s)
The driving period exactly matches the natural period. Resonance condition is satisfied.
$f_{drive} = 1/2.0 = 0.5\text{ Hz} = f_0$
Step 4
Conclusion
In Case 2, resonance occurs — amplitude builds up enormously. In Case 1, the off-resonance driving produces only small oscillations.
$f_{drive} = f_0 \implies \text{resonance (max amplitude)}$
Memorize formulas
Total Energy (Spring)
Maximum energy stored in a spring-mass system.
Formula
$E_{total} = \frac{1}{2}kA^2$
Total Energy (General SHM)
Energy using mass and angular frequency.
Formula
$E_{total} = \frac{1}{2}m\omega^2A^2$
Kinetic Energy at Position x
Kinetic energy as function of displacement.
Formula
$K.E. = \frac{1}{2}k(A^2 - x^2)$
Potential Energy at Position x
Elastic potential energy at displacement x.
Formula
$P.E. = \frac{1}{2}kx^2$
Velocity at Position x
Speed of oscillator at any displacement.
Formula
$v = \omega\sqrt{A^2 - x^2}$
Resonance Condition
Maximum amplitude when driving matches natural frequency.
Formula
$f_{driving} = f_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
Review
1
An object undergoing SHM has a total energy $E$. When the displacement is half the amplitude ($x = A/2$), what is the ratio of Kinetic Energy to Total Energy?
Ans
$3/4$
Review
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Find the potential-energy fraction at half amplitude
For SHM, potential energy varies as $x^2$. At $x = A/2$, the potential energy is one quarter of the total energy at amplitude.
$$U = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{4}\left(\frac{1}{2}kA^2\right) = \frac{E}{4}$$
Subtract from total energy
The rest of the energy is kinetic, so take total energy minus potential energy.
$$K = E - \frac{E}{4} = \frac{3E}{4} \Rightarrow \frac{K}{E} = \frac{3}{4}$$
2
A 0.80 kg block on a frictionless surface is attached to a spring ($k = 500$ N/m) and pulled 4.0 cm from equilibrium. What is the block's speed when it is 2.0 cm from equilibrium?
Ans
0.87 m/s
Review
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Apply conservation of mechanical energy
At the pulled-back position, the spring stores total energy. At $x = 2.0$ cm, part of that energy is still potential and the rest is kinetic.
$$\frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$$
Solve for speed and substitute values
Rearrange for $v$ and use meters for the displacements.
$$v = \sqrt{\frac{k}{m}(A^2 - x^2)} = \sqrt{\frac{500}{0.80}(0.040^2 - 0.020^2)} = \sqrt{0.75} \approx 0.87\ \text{m/s}$$
3
A driving force is applied to an oscillating system. As the damping in the system increases, what happens to the resonance peak?
Ans
Peak height decreases and the curve becomes wider
Review
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Connect damping to energy loss
Greater damping means more energy is removed each cycle, so the oscillator cannot build up as large an amplitude at resonance.
Peak height decreases
Recognize the shape of the resonance curve
A stronger damping effect also makes the response less selective in frequency, so the resonance curve spreads out.
The curve becomes wider
4
The total energy of a simple harmonic oscillator is $20$ J. If the mass is doubled and the frequency is halved while keeping the amplitude the same, what is the new total energy?
Ans
$10$ J
Review
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Use the SHM energy dependence
For fixed amplitude, total energy scales as $E \propto mf^2$ because $\omega = 2\pi f$.
$$E \propto mf^2$$
Apply the new mass and frequency
Doubling the mass multiplies energy by 2, while halving the frequency multiplies it by $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.
$$E' = 2 \times \frac{1}{4} \times 20 = 10\ \text{J}$$
5
A 2.0 kg block sliding at 3.0 m/s on a frictionless surface hits a horizontal spring ($k = 1800$ N/m) and compresses it. What is the maximum compression of the spring?
Ans
0.10 m
Review
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Set kinetic energy equal to spring energy
At maximum compression, the block momentarily stops, so all of its initial kinetic energy is stored in the spring.
$$\frac{1}{2}mv^2 = \frac{1}{2}kd^2$$
Solve for compression
Rearrange for $d$ and substitute the given values.
$$d = v\sqrt{\frac{m}{k}} = 3.0\sqrt{\frac{2.0}{1800}} \approx 0.10\ \text{m}$$
6
An SHM oscillator has total energy $E$. At what displacement is the potential energy three times the kinetic energy?
Ans
$x = A\sqrt{3}/2$
Review
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Convert the energy ratio into fractions of total energy
If potential energy is three times kinetic energy, then the total energy is split as 3 parts potential and 1 part kinetic.
$$U = \frac{3E}{4}, \qquad K = \frac{E}{4}$$
Use the SHM potential-energy formula
Since $U = \frac{1}{2}kx^2$ and $E = \frac{1}{2}kA^2$, compare the two expressions and solve for $x$.
$$\frac{x^2}{A^2} = \frac{3}{4} \Rightarrow x = \frac{A\sqrt{3}}{2}$$
7
A 0.50 kg mass on a spring ($k = 200$ N/m) oscillates with an amplitude of 3.0 cm. What is the maximum speed of the mass?
Ans
0.60 m/s
Review
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Find the angular frequency
For a mass-spring system, $\omega = \sqrt{k/m}$.
$$\omega = \sqrt{\frac{200}{0.50}} = \sqrt{400} = 20\ \text{rad/s}$$
Use the SHM speed relation
The maximum speed occurs at equilibrium, where $v_{max} = A\omega$.
$$v_{max} = 0.030 \times 20 = 0.60\ \text{m/s}$$
8
A damped oscillator starts with amplitude 12 cm and total energy 36 J. After some time, its amplitude is 6.0 cm. How much energy has been lost to damping?
Ans
27 J
Review
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Use the amplitude-squared energy rule
In SHM, total energy varies as the square of amplitude, so halving amplitude leaves one quarter of the energy.
$$\frac{E_2}{E_1} = \left(\frac{6.0}{12}\right)^2 = \frac{1}{4}$$
Compute the lost energy
Find the remaining energy first, then subtract it from the initial energy.
$$E_2 = \frac{1}{4}(36) = 9.0\ \text{J}, \qquad E_{\text{lost}} = 36 - 9.0 = 27\ \text{J}$$
9
Which of the following is an example of resonance?
Ans
A radio receiver tuned to pick up a specific broadcast frequency
Review
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Identify the resonance condition
Resonance occurs when a system is driven at or near its natural frequency, producing a much larger response.
Natural frequency matches driving frequency
Match the option to the definition
A tuned radio circuit is designed to select one broadcast frequency, so it is the clearest example of resonance.
A radio receiver tuned to a specific broadcast frequency
10
A 900 kg car is supported by four identical springs. When a 75 kg driver sits in the car, the chassis drops 1.0 cm. What is the natural frequency of vertical vibration with the driver?
Ans
1.37 Hz
Review
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Use the static drop to find the effective spring constant
The driver's weight causes a 1.0 cm compression, so the combined spring constant follows from $mg = k_{eff}\Delta x$.
$$k_{eff} = \frac{mg}{\Delta x} = \frac{75 \times 9.8}{0.010} = 73500\ \text{N/m}$$
Compute the vibration frequency from $f = \frac{1}{2\pi}\sqrt{k/m}$
Use the total vibrating mass of the car plus driver.
$$f = \frac{1}{2\pi}\sqrt{\frac{73500}{975}} \approx 1.37\ \text{Hz}$$
Question 1 of 10
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1
An object undergoing SHM has a total energy $E$. When the displacement is half the amplitude ($x = A/2$), what is the ratio of Kinetic Energy to Total Energy?