Energy and the Simple Harmonic Oscillator

Saul Beceiro-Novo

Oscillatory Motion, Waves, and the Physics of Hearing

120 Energy and the Simple Harmonic Oscillator

Learning Objectives

Determine the maximum speed of an oscillating system.

To analyze the energy in a simple harmonic motion system, we begin by identifying all the energy forms it can possess. From Hooke’s Law: Stress and Strain Revisited, we know that a stretched or compressed spring stores elastic potential energy, expressed as:

[latex]{\text{PE}}_{\text{el}} = \frac{1}{2}kx^2[/latex]

In the absence of energy loss (e.g., no friction), total mechanical energy is conserved. That means the total of kinetic energy and potential energy remains constant:

[latex]\text{KE} + {\text{PE}}_{\text{el}} = \text{constant}[/latex]

Or, in terms of variables:

[latex]\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}[/latex]

This applies to any SHM system, including pendulums. For a simple pendulum, we use the substitutions:

Velocity: [latex]v = L\omega[/latex]
Spring constant: [latex]k = \frac{mg}{L}[/latex]
Displacement: [latex]x = L\theta[/latex]

This gives us an energy expression for pendulums:

[latex]\frac{1}{2}mL^2\omega^2 + \frac{1}{2}mgL\theta^2 = \text{constant}[/latex]

In an undamped SHM system, energy oscillates between kinetic and potential forms. For example, in a spring-mass system on a frictionless surface, as illustrated in Figure 120.1, the energy shifts from entirely potential (stored in the spring) to entirely kinetic (when passing equilibrium) and back again.

Figure a shows a spring on a frictionless surface attached to a bar or wall from the left side, and on the right side of it there’s an object attached to it with mass m, its amplitude is given by X, and x equal to zero at the equilibrium level. Force F is applied to it from the right side, shown with left direction pointed red arrow and velocity v is equal to zero. A direction point showing the north and west direction is also given alongside this figure as well as with other four figures. The energy given here for the object is given according to the velocity. In figure b, after the force has been applied, the object moves to the left compressing the spring a bit, and the displaced area of the object from its initial point is shown in sketched dots. F is equal to zero and the V is max in negative direction. The energy given here for the object is given according to the velocity. In figure c, the spring has been compressed to the maximum level, and the amplitude is negative x. Now the direction of force changes to the rightward direction, shown with right direction pointed red arrow and the velocity v zero. The energy given here for the object is given according to the velocity. In figure d, the spring is shown released from the compressed level and the object has moved toward the right side up to the equilibrium level. F is zero, and the velocity v is maximum. The energy given here for the object is given according to the velocity. In figure e, the spring has been stretched loose to the maximum level and the object has moved to the far right. Now again the velocity here is equal to zero and the direction of force again is to the left hand side, shown here as F is equal to zero. The energy given here for the object is given according to the velocity. — Figure 120.1: Energy transformation during SHM of a spring-mass system. Energy transitions between elastic potential energy and kinetic energy as the object oscillates.

We can derive the velocity of an object in SHM using conservation of energy. Suppose the object starts from rest at its maximum displacement [latex]x = X[/latex], then the total energy is:

[latex]E = \frac{1}{2}kX^2[/latex]

At any point in the motion:

[latex]\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kX^2[/latex]

Solving for [latex]v[/latex], we get:

[latex]v = \pm \sqrt{\frac{k}{m}(X^2 - x^2)}[/latex]

This can be rewritten as:

[latex]v = \pm \sqrt{\frac{k}{m}}X \sqrt{1 - \frac{x^2}{X^2}}[/latex]

Or in terms of maximum velocity:

[latex]v = \pm v_{\text{max}} \sqrt{1 - \frac{x^2}{X^2}}[/latex]

Where:

[latex]v_{\text{max}} = \sqrt{\frac{k}{m}}X[/latex]

Maximum velocity occurs at the equilibrium position ([latex]x = 0[/latex]). This velocity depends on:

Amplitude: Larger amplitude [latex]X[/latex] increases [latex]v_{\text{max}}[/latex].
Spring stiffness: Greater spring constant [latex]k[/latex] leads to higher [latex]v_{\text{max}}[/latex].
Mass: Heavier objects ([latex]m[/latex]) reduce [latex]v_{\text{max}}[/latex] because acceleration is lower.

For a simple pendulum, we can derive a similar expression using angular displacement:

[latex]\omega_{\text{max}} = \sqrt{\frac{g}{L}} \, \theta_{\text{max}}[/latex]

This relates the pendulum’s angular velocity to its maximum angular displacement [latex]\theta_{\text{max}}[/latex], length [latex]L[/latex], and gravitational acceleration [latex]g[/latex].

Example 120.1: Determine the Maximum Speed of an Oscillating System: A Bumpy Road

Suppose that a car is 900 kg and has a suspension system that has a force constant [latex]k=6\text{.}\text{53}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N/m}[/latex]. The car hits a bump and bounces with an amplitude of 0.100 m. What is its maximum vertical velocity if you assume no damping occurs?

Strategy

We can use the expression for [latex]{v}_{\text{max}}[/latex] given in [latex]{v}_{\text{max}}=\sqrt{\frac{k}{m}}X[/latex] to determine the maximum vertical velocity. The variables [latex]m[/latex] and [latex]k[/latex] are given in the problem statement, and the maximum displacement [latex]X[/latex] is 0.100 m.

Solution

Identify known values.
Substitute known values into [latex]{v}_{\text{max}}=\sqrt{\frac{k}{m}}X[/latex]:
[latex]{v}_{\text{max}}=\sqrt{\frac{6\text{.}\text{53}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N/m}}{\text{900}\phantom{\rule{0.25em}{0ex}}\text{kg}}}(0\text{.}\text{100}\phantom{\rule{0.25em}{0ex}}\text{m)}[/latex]
Calculate to find [latex]{v}_{\text{max}}=\text{0.852 m/s}.[/latex]

Discussion

This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find [latex]{v}_{\text{max}}[/latex]. We could use it directly, as was done in the first example featured in Hooke’s Law: Stress and Strain Revisited.

The small vertical displacement [latex]y[/latex] of an oscillating simple pendulum, starting from its equilibrium position, is given as

[latex]y\left(t\right)=a\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\omega t}[/latex]

where [latex]a[/latex] is the amplitude, [latex]\omega[/latex] is the angular velocity and [latex]t[/latex] is the time taken. Substituting [latex]\omega =\frac{2\pi }{T}[/latex], we have

[latex]yt=a\phantom{\rule{0.25em}{0ex}}\text{sin}\left(\frac{2\pi t}{T}\right)[/latex]

Thus, the displacement of the pendulum is a function of time as shown above. Also, the velocity of the pendulum is given by

[latex]v\left(t\right)=\frac{2\mathrm{a\pi }}{T}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\frac{2\pi t}{T}\right)[/latex]

so the motion of the pendulum is a function of time.

Check Your Understanding

Why does it hurt more if your hand is snapped with a ruler than with a loose spring, even if the displacement of each system is equal?

Answer: The ruler is a stiffer system. A stiffer system has a higher spring constant [latex]k[/latex], which means it exerts more force for the same displacement. More force results in a greater impact, which causes more pain.

You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system.

Answer: You could increase the mass of the object. Since the maximum velocity is inversely related to the square root of the mass, a heavier object will oscillate more slowly.

Section Summary

In an undamped SHM system, energy is continuously exchanged between kinetic and potential forms, while the total mechanical energy remains constant:
[latex]\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}[/latex]
The maximum velocity of an oscillator depends on:
- Amplitude [latex]X[/latex] (larger amplitude means higher velocity),
- Spring constant [latex]k[/latex] (stiffer springs give higher force and speed),
- Mass [latex]m[/latex] (heavier masses move slower).
The maximum velocity is given by:

[latex]v_{\text{max}} = \sqrt{\frac{k}{m}}X[/latex]

Healthcare Connection

Because the energy of a simple harmonic oscillator is proportional to the square of its amplitude, doubling the amplitude of a vibration quadruples the energy it carries, not just doubles it. This matters for hearing health: a sound wave with twice the amplitude delivers four times the energy to the delicate hair cells of the inner ear, which is part of why loud sounds are so disproportionately damaging. Repeated exposure to high-amplitude sound waves, from concerts, power tools, or headphones turned up too high, can deliver enough mechanical energy to permanently damage these hair cells, which is a primary cause of noise-induced hearing loss and why hearing protection becomes increasingly important as sound amplitude increases.

Conceptual Questions

Explain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Also explain how a driving mechanism can compensate. (A pendulum clock is such a system.)

Problems & Exercises

The length of nylon rope from which a mountain climber is suspended has a force constant of [latex]1\text{.}\text{40}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N/m}[/latex].
1. What is the frequency at which he bounces, given his mass plus the mass of his equipment is equal to 90.0 kg?
2. How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack?
3. Repeat both parts of this problem in a situation where twice this length of nylon rope is used.
Near the top of the Citigroup Center building in New York City, there is an object with a mass of [latex]4\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex] on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building—the driving force is transferred to the object, which oscillates instead of the entire building.
1. What effective force constant should the springs have to make the object oscillate with a period of 2.00 s?
2. What energy is stored in the springs for a 2.00-m displacement from equilibrium?

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