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3.1 Sinusoidal excitation of a resonator
In this series of postings we will discuss sinusoidal excitation of a resonator, specifically directed towards mass-and-spring resonators. We chose this resonator because it makes for very visual animations. The math is the same for all simple resonators. Immediately below are animations of variations of the the mass and spring resonator. Mouse over each one to see the action. We will work with the variation shown in Fig. 1b, a mass hanging from a spring which is attached to a rigid support.
Fig. 1b. Animation of the particular mass/spring resonator discussed next.
The mechanical resonator
Figure 2 shows the simple mass and spring resonator. We will now derive an equation for motion of this resonator.
The spring exerts a force on the mass of:
Fspring = −kx , (1)
where k is the spring constant of the spring and x is the vertical displacement of the mass from its equilibrium position. Springs for which k is constant and does not vary with x are called "linear" because the graph of force versus x is a straight line as shown in Fig. 3. A stiff, i.e. strong, spring has a larg spring constant while a weak spring has a small spring constant.
Fig. 3. Force generated by a linear spring versus the distance the spring is stretched or compressed.
We combine (1) with Newton's second law of motion which governs the motion of all objects moving slowly compared to the speed of light. In this case we are considering the effect that the spring force has on the motion of the mass (i.e. the bob):
This equation relates the force on the bob to the acceleration the bob will experience because of the force. In (2) m is the mass of the bob and t is time. The operator d2/dt2 performs a second time derivative on the displacement x which is a function of time. This second derivative operation yields the acceleration. The displacement x is the vertical distance the bob is displaced from its position when the spring is unstretched and uncompressed.
Effect of gravity
In addition to the spring, gravity also exerts a force on the bob. Gravity exerts a constant downward force of:
Fgravity = −mg , (3)
where g = 9.8 m/s2 is the local acceleration of gravity for the surface of the Earth.
We add the force of the spring (equation 2) to that of gravity (equation 4) and substitute this total into (2) to have:
In general gravity will make the bob move down a little. Anticipating this, we assume that x will be of the form:
x = x0 + x1(t) , (5)
where x0 is a constant shift (or sag) due to gravity and x1(t) is the time varying part of the displacement. Substituting (5) into (4), we get:
Considering the case where the time varying part is zero, (6) becomes −kx0 − mg = 0. This can be solved for x0, i.e.: x0 = −mg/k . This gives the static "sag" of the spring due to gravity. Substituting this value for x0 into (6) gives:
Equation 8 is basically the same as equation 5 without the complication of gravity. Another difference is that the time varying part x1 is the displacement of the bob from the "sagging" position. If we remember this modified displacement definition, then we are free to just ignore gravity. In this spirit, we rewrite (7) just using this new definition of x:
I might note that we could not separate out the effect of gravity so simply if the spring were not linear (as graphed in Fig. 3 above).
Fig. 4. Motion of a mass/string resonator as recorded by a strip chart recorder (similar to those used in EKG machines, seismographs and polygraphs). The particular solution shown is x = A sin ωt. Note that on this type of recorder, time progresses towards the left, backwards of the normal convention. Mouse over the figure to see the action.
Solutions to the differential equation
We can rearrange (8) to be in the standard second order differential equation format as:
The standard solutions to the very simple second order differential equation as given by (9) are sines, cosines, and complex exponentials in the forms:
x = A sin ωt, x = A cos ωt, x = A e±iωt (10)
and sums of these. All of these mathematically describe the oscillations that we associate with a mass dangling at the end of a spring.
Fig. 5. This animation shows how various parameters of this resonator vary versus time. The red arrow on the bob indicates the spring force on the bob, while the blue arrow indicates the bob's momentum. For discussion of the interactions of force, momentum, and energy see my earlier posting. Note that the total energy of the resonator is constant but varies in its form and location, between the form of potential energy in the spring and kinetic energy in the mass. Mouse over the figure to see the action. Click on it to restart.
The angular frequency ω (in radians per second) and regular frequency f (in Hz) of the oscillations are given by:
Equation 12 for ω can be had by substituting any of the standard solutions (10) back into (9) and solving for ω. The frequency f is related to ω by the definition of ω, i.e. by ω = 2πf. The amplitude A of the oscillation is arbitrary and is determined by the impulse that is given to the mass to set it into oscillations.
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