To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke’s Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by:
Because a simple harmonic oscillator has no dissipative forces, the other important form of energy is kinetic energy $\text{KE}$ . Conservation of energy for these two forms is:
or
This statement of conservation of energy is valid for all simple harmonic oscillators, including ones where the gravitational force plays a role
Namely, for a simple pendulum we replace the velocity with $v=L\omega$ , the spring constant with $k=mg/L$ , and the displacement term with $x=L\theta$ . Thus
In the case of undamped simple harmonic motion, the energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, as shown again in [Figure 1], the motion starts with all of the energy stored in the spring. As the object starts to move, the elastic potential energy is converted to kinetic energy, becoming entirely kinetic energy at the equilibrium position. It is then converted back into elastic potential energy by the spring, the velocity becomes zero when the kinetic energy is completely converted, and so on. This concept provides extra insight here and in later applications of simple harmonic motion, such as alternating current circuits.
The conservation of energy principle can be used to derive an expression for velocity $v$ . If we start our simple harmonic motion with zero velocity and maximum displacement ( $x=X$ ), then the total energy is
This total energy is constant and is shifted back and forth between kinetic energy and potential energy, at most times being shared by each. The conservation of energy for this system in equation form is thus:
Solving this equation for $v$ yields:
Manipulating this expression algebraically gives:
and so
where
From this expression, we see that the velocity is a maximum ( ${v}_{\text{max}}$ ) at $x=0$ , as stated earlier in $v\left(t\right)=-{v}_{\text{max}} \sin \frac{2\pi t}{T}$ . . Notice that the maximum velocity depends on three factors. Maximum velocity is directly proportional to amplitude. As you might guess, the greater the maximum displacement the greater the maximum velocity. Maximum velocity is also greater for stiffer systems, because they exert greater force for the same displacement. This observation is seen in the expression for ${v}_{\text{max}};$ it is proportional to the square root of the force constant $k$ . Finally, the maximum velocity is smaller for objects that have larger masses, because the maximum velocity is inversely proportional to the square root of $m$ . For a given force, objects that have large masses accelerate more slowly.
A similar calculation for the simple pendulum produces a similar result, namely:
Suppose that a car is 900 kg and has a suspension system that has a force constant $k= 6.53 \times 10^{4} \text{N/m}$ . 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 ${v}_{\text{max}}$ given in ${v}_{\text{max}}=\sqrt{\frac{k}{m}}X$ to determine the maximum vertical velocity. The variables $m$ and $k$ are given in the problem statement, and the maximum displacement $X$ is 0.100 m.
Solution
Substitute known values into ${v}_{\text{max}}=\sqrt{\frac{k}{m}}X$
:
Discussion
This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find ${v}_{\text{max}}$ . We could use it directly, as was done in the example featured in Hooke’s Law: Stress and Strain Revisited.
The small vertical displacement $y$ of an oscillating simple pendulum, starting from its equilibrium position, is given as
where $a$ is the amplitude, $\omega$ is the angular velocity and $t$ is the time taken. Substituting $\omega =\frac{2\pi }{T}$ , we have
Thus, the displacement of pendulum is a function of time as shown above.
Also the velocity of the pendulum is given by
so the motion of the pendulum is a function of time.
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?
The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater force, which hurts more.
You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system.
You could increase the mass of the object that is oscillating.
Energy in the simple harmonic oscillator is shared between elastic potential energy and kinetic energy, with the total being constant:
Maximum velocity depends on three factors: it is directly proportional to amplitude, it is greater for stiffer systems, and it is smaller for objects that have larger masses:
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.)
The length of nylon rope from which a mountain climber is suspended has a force constant of $1.40 \times 10^{4} \text{N/m}$.
(a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg?
(b) 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? Hint: Use conservation of energy.
(c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.
Strategy
For part (a), the frequency of oscillation for a mass-spring system is $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$. For part (b), we use conservation of energy: the gravitational potential energy lost during the fall equals the elastic potential energy stored in the rope plus the gravitational potential energy lost while the rope stretches. For part (c), when the rope length doubles, the spring constant is halved (springs in series).
Solution
(a) Using the frequency formula for a mass-spring system:
(b) Using conservation of energy. The climber falls distance $h = 2.00$ m, then the rope stretches distance $x$. Taking the lowest point as the reference level:
Initial energy (at top of fall): $E_i = mg(h + x)$
Final energy (at bottom): $E_f = \frac{1}{2}kx^2$
Substituting values:
Using the quadratic formula:
(c) When the rope length doubles, the effective spring constant becomes $k' = \frac{k}{2} = 7.00 \times 10^{3}$ N/m.
For frequency:
For the stretch, using the same energy conservation approach with $k' = 7.00 \times 10^{3}$ N/m:
Discussion
The frequency of 1.99 Hz for the climber bouncing on the rope is reasonable—about 2 oscillations per second. This is slow enough to be noticeable but fast enough to be uncomfortable.
For the fall arrest, the rope stretches 50.2 cm (about half a meter), which seems reasonable for a nylon rope with this spring constant catching a 2-meter fall. The energy of the fall is absorbed by stretching the rope elastically.
When the rope length doubles, the spring constant halves because the same force is distributed over twice the length. This makes the rope “softer”—the frequency decreases by a factor of $\sqrt{2}$ to 1.41 Hz, and the rope stretches more (71.0 cm instead of 50.2 cm) when catching the same fall. The longer rope is actually safer for fall arrest because it stretches more, reducing the peak deceleration force on the climber.
Answer
(a) 1.99 Hz
(b) 50.2 cm
(c) 1.41 Hz, 0.710 m
Engineering Application
Near the top of the Citigroup Center building in New York City, there is an object with mass of $4.00 \times 10^{5} \text{kg}$ 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 is being driven—the driving force is transferred to the object, which oscillates instead of the entire building. (a) What effective force constant should the springs have to make the object oscillate with a period of 2.00 s? (b) What energy is stored in the springs for a 2.00-m displacement from equilibrium?
Strategy
For part (a), we use the relationship between period and spring constant for a mass-spring oscillator: $T = 2\pi\sqrt{\frac{m}{k}}$. We solve for $k$. For part (b), the elastic potential energy stored in the springs is $PE_{el} = \frac{1}{2}kx^2$, where $x$ is the displacement from equilibrium.
Solution
(a) Starting with the period equation:
Square both sides:
Solve for $k$:
Substitute the given values:
(b) The elastic potential energy for a displacement of $x = 2.00$ m:
Discussion
This tuned mass damper system is an elegant engineering solution to reduce building sway. The spring constant of $3.95 \times 10^{6}$ N/m is enormous—about 4 million newtons per meter—which is necessary to control the motion of the 400-ton mass with a 2-second period.
The stored energy of 7.90 MJ (megajoules) during a 2-meter displacement is substantial, equivalent to the kinetic energy of a large truck moving at highway speed. This energy represents the oscillation energy that would otherwise be causing the building to sway uncomfortably.
The key to this system’s effectiveness is resonance: by tuning the mass damper to oscillate at the same frequency as the building’s natural oscillation (2.00 s period), the damper absorbs energy from the building’s motion. When wind forces try to make the building sway, the mass damper oscillates out of phase with the building, creating forces that oppose and reduce the building’s motion. This is why the spring constant must be adjustable—it allows engineers to fine-tune the system as building conditions change.
Answer
(a) $3.95 \times 10^{6} \text{ N/m}$
(b) $7.90 \times 10^{6} \text{ J}$