Energy and the Simple Harmonic Oscillator
By the end of this section, you will be able to:
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 \(\mathrm{KE}\). 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.
Figure 1 The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface.
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:
where
From this expression, we see that the velocity is a maximum \((v_{\mathrm{max}}) \) at \(x =0\), as stated earlier in \(v(t) = -v_{\mathrm{max}} \sin \frac{2 \pi t}{T}\). 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 vmax;vmax; 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:
Consider a mass m attached to a spring, with spring constant k, fixed to a wall. When the mass is displaced from its equilibrium position and released, the mass undergoes simple harmonic motion. The spring exerts a force \(F = - kv\) on the mass. The potential energy of the system is stored in the spring. It will be zero when the spring is in the equilibrium position. All the internal energy exists in the form of kinetic energy, given by \(KE = \frac{1}{2} mv^2\). As the system oscillates, which means that the spring compresses and expands, there is a change in the structure of the system and a corresponding change in its internal energy. Its kinetic energy is converted to potential energy and vice versa. This occurs at an equal rate, which means that a loss of kinetic energy yields a gain in potential energy, thus preserving the work-energy theorem and the law of conservation of energy.
Suppose that a car is 900 kg and has a suspension system that has a force constant \(k= 6.53 \times 10^4 \thinspace \mathrm{N/m}\). 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 \(v_{\mathrm{max}}\) given in \(v_{\mathrm{max}} = \sqrt{\frac{k}{m}}X\) 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
Discussion
This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find \(v_{\mathrm{max}}\). 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.
You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system.
What is the total energy of a 6 kg block tied to a spring with a 19 \(N\over m\) constant? The amplitude of the movement is 1 m.
A 6 block attached to a spring has an energy of 6 J when it reaches its maximum displacement. If the spring constant to which is attached is 27 \(N \over m\), what is the maximum displacement, in meters, of the block?
A 9 kg block rests on a frictionless surface and is connected to a spring whose restitution constant is 112 \(N\over m\). If the block moves 42 cm from its equilibrium position, what maximum speed does the block reach? Express your answer in meters per second.
The length of nylon rope from which a mountain climber is suspended has a force constant of \(1.40\times 10^4\; N/m\). What is the frequency at which he bounces, given his mass plus and the mass of his equipment is 90.0 kg?
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. Ignore the energy the climber gains as the rope stretches.
Express your answer in centimeters.
Which of the following describes the simple harmonic motion in a horizontal-spring block oscillator system?
There is a system composed of 4 kg mass attached to a spring whose constant is 84 \(N\over m\) and can reach a maximum displacement of 22 m. How fast is it if has reached 71\(\%\) of its maximum displacement? Express your answer in meters per second.
Near the top of the Citigroup Center building in New York City, there is an object with a mass of \(4.00\times10^5\;kg\) 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. What effective force constant should the springs have to make the object oscillate with a period of 2.00 s?
What energy is stored in the springs for a 2.00-m displacement from equilibrium?
This is a large modal.
