Chapter One Introduction to Oscillations This chapter discusses simple oscillations. This chapter also discusses simple functions to model oscillations like the sine and cosine functions. Part One Sine and Cosine Functions We will find that simple oscillations can be modeled mathematically with sine and cosine functions. Consider the following function: x= Asin Next we'll add the condition that the angle ( Θ ) is changing at a constant rate. Therefore we can say that the angular frequency ( ω ), which is the change in the angle per unit time, is constant. = t = t x= Asin t Here the independent variable is t for time, and the dependent variable is x for position. The amplitude of the function is labeled as A. The amplitude is absolute value of the maximum x, the height of the graph. The variable ω is the angular frequency. We will consider the amplitude and angular frequency to be a constant. Let's put in some numbers and look at a plot of this function. x t =5msin 4 s 1 t 5 Position vs. Time 4 3 2 x (meters) 1 0-1 -2-3 -4-5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 t (seconds) Here the amplitude is 5 meters and the angular frequency is 2 s -1.
You will notice that the maximum of the function is 5.0 m and the minimum of the function is -5.0 m. We will define the amplitude as the absolute magnitude of the maximum of the dependent variable (x), so the amplitude of the function is 5.0 m. 5 Position vs. Time 4 3 2 A x (meters) 1 0-1 -2-3 -4-5 A 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 t (seconds) The period is the length of time required for the function to repeat. The period (T) of the function shown is 0.5 seconds. Recall the angular frequency is 2 π radians divided by the period, so the period is 2 π radians divided by the angular frequency. = 2 T T = 2 T = 2 4 s 1 T =0.5 s
x (meters) 5 4 3 2 1 0-1 -2-3 Position vs. Time T -4-5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 t (seconds) EXERCISE ONE: In this exercise, you will plot a cosine function. Consider the function: x = A cos t =0.2 m cos 1.57 s 1 t 1. Create a spreadsheet. a) Use the Create Names or Define Names function to define the constants for Amplitude and Angular Frequency. b) Create a column labeled as Time(s) and populate it from 0.0 seconds to 8.0 seconds. c) Create a column labeled as x(m) and populate it with the above equation. 2. Plot x(m) versus time(s). 3. Calculate the period and the frequency of the function. Does your plot mirror your calculations? 4. Create a column labeled as v ave (m/s). Populate this column with the numerical v difference which we will call v ave : ave = x t = x i 1 x i 1 t i 1 t i 1 5. Plot v ave (m/s) versus t(s). What function does this look like? ** See the Appendix A if you need help with spread sheeting mechanics. *** Hand in a copy of each plot with a cover sheet with your name and the answers to any questions asked. Staple them together.
Part Two ass on a Spring Consider a simple perfect spring with a spring constant (k), laying flat on a frictionless table and attached to the wall. x=0.0 m Next, let's attach a mass to the spring. When the spring is at rest, we will mark the free end of the spring at x = 0.0m. If we displace the mass in the negative x direction, back to x = -A, then the spring pushes back on the mass in the positive x direction with a magnitude of F s =kx=ka. F s x=-a x=0.0 m If we pull the mass to x = +A, then the spring pulls back in the negative x direction with a magnitude of F s =kx=ka. F s x=-a x=0.0 m x=+a Now if we pull the mass out to x = +A, hold, and then release, the mass will oscillate back and forth between x = A and x = -A. Note that when the mass is at x = +A or x = -A, the velocity is equal to zero. Also, when the mass is at x = +A and x = -A, the force provided by the spring is at a maximum value. Since the force is maximum, so is the
acceleration. When we release the mass at x = +A, it starts from rest. The spring pulls the mass in the negative x direction and the velocity increases in the negative x direction. The mass continues to speed up until it reaches the x = 0.0 m position. At that point the force provided by the spring is equal to zero. Then the mass continues to move in the negative x direction, and compresses the spring. As the mass does this, the spring responds with a force in the positive direction. The mass continues to slow down until it stops at x = -A. At this point the spring continues to provide a force in the positive x direction. The mass begins to increase its velocity in the positive x direction. It continues to increase until the mass's velocity reaches its maximum value at x = 0.0m. This is where the force of the spring equals zero. At this point, the mass continues in the positive x direction, stretching the spring. The spring responds by producing a force in the negative x direction. The effect of this force is to slow the mass until it stops at x = +A. At this point, the force provided by the spring has a maximum value in the negative x direction. The mass then begin to repeat its motion. It continues to oscillate between x = +A and x = -A. Since there is not friction between the table and the mass, there is no energy used doing work against friction, i.e there is no energy lost. Since we are considering an ideal spring, there is no energy lost in the spring. The mass will continue to oscillate forever. This motion is called Simple Harmonic otion.
A Detailed Look of the otion of a ass on a Spring x=-a x=0.0 m x=+a The mass starts at x=+a with a velocity of zero, where the spring exerts a force in the negative x direction. Here the magnitude of the force is maximum and equal to F=-kA. The mass begins to accelerate in the negative x direction since the spring exerts a force in the negative x direction. Here the magnitude of the force is decreasing and is equal to F=-kx The velocity of the mass increases in the negative x direction. The mass continues to accelerate in the negative x direction until it reaches its maximum value at x=0.0m. Here the force provided by the spring is equal to zero because the spring is neither stretched or compressed. The mass continues to move in the negative x direction, but the force provided by the spring is now in the positive x direction. The force is equal to F=+kx. The acceleration is therefore in the positive x direction and the velocity is still in the negative x direction, but decreasing in magnitude. The mass continues to move in the negative x direction, with the force provided by the spring in the positive x direction. The force is equal to F=+kx and the magnitude is increasing. The acceleration is in the positive x direction and the velocity decreases to zero. The force reaches a maximum value of F=+kA at x=a. The mass stops, turns around and continues to move in the positive x direction. After the mass stops and turns around at x=a, the compressed spring provides a force in the positive x direction and has a magnitude equal to F=+kA. An acceleration in the positive x direction is exerted on the mass. The velocity of the mass is now increasing as it moves in the positive x direction.
The mass continues to speed up, moving in the positive x direction, until it reaches a maximum value at x=0.0m. At this point the force provided by the spring is equal to zero. The mass continues to move in the positive x direction, but the force provided by the spring is now exerted in the negative x direction. It is equal to F=-kx. The velocity is decreasing and the mass is slowing down. The mass continues to move in the positive x direction, until the velocity slows to zero at x=a. The force has reached a maximum value and is equal to F=-kA. The mass stops, turns around, and continues to move in the negative x direction. Now the mass starts to move in the negative x direction, starts the oscillations all over again.
Below shows a plot of position versus time for a mass on an ideal spring oscillating between x = +0.2 m and x = -0.2 m. ass on a Spring Position vs. Time x(m) 0.30 0.25 0.20 0.15 0.10 0.05 0.00-0.05-0.10-0.15-0.20-0.25-0.30 0 1 2 3 4 5 6 7 8 9 10 11 12 t(s) It can be seen that the amplitude (A) is 0.2 m. You can also read the period (T) is 4.0 s. This means that the angular frequency is = 2 T = 2 4 s = 1.57 s 1. Looking at the plot, it looks like the position of the mass as a function of time can be modeled as a cosine function. x t = A cos t In the above plot: x t = A cos t = 0.2 m cos 1.57 s 1 t
Next, let's look at the velocity. As you recall velocity at some time (t) is equal to the change in position with respect to time. This is the slope of a line tangential to a point on a position versus time plot. v t = lim t 0 x t = lim t 0 x i 1 x i 1 t i 1 t i 1 Look at the plot of position versus time. Notice that the places were the slope of the this graph is equal to zero correspond to the places were the position x is equal to the Amplitude of the motion (A). x(m) 0.30 0.25 0.20 0.15 0.10 0.05 0.00-0.05-0.10-0.15-0.20-0.25 ass on a Spring Position vs. Time -0.30 0 1 2 3 4 5 6 7 8 9 10 11 12 This fits the description of the motion of the mass on the spring. Next, let's zoom in on one period of the plot. We will look the slope of the graph at a couple of points and see if this fits the description of the motion. t(s)
x(m) 0.30 0.25 0.20 0.15 0.10 0.05 0.00-0.05-0.10-0.15-0.20-0.25-0.30 A B ass on a Spring Position vs. Time C D 0 0.5 1 1.5 2 2.5 3 3.5 4 t(s) Notice that the slope at point B has a larger magnitude than the slope at point A. The slope at point C has a larger magnitude than the slope at point B. Then the magnitude of the slope appears to decrease as we look at point D. Recall that the slope of these lines are related to the velocity, so the maximum magnitude of the velocity is attained at x = 0.0 m. If we continue this analysis, we see that the magnitude of the velocity continue to decrease until the mass stops at x = A and then begins to increase in magnitude until it reaches a maximum at x = 0.0 m. The magnitude of the velocity then decreases until the mass stops at x = A, were the mass then turns around and start it's oscillation all over again. A plot of velocity (v) versus time (t) will look like the plot below. Notice that this plot has the form of a negative sine plot. We can model the velocity as a negative sine plot.
v(m/s) 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0-0.05-0.1-0.15-0.2-0.25-0.3-0.35 v t = v max sin t ass on a Spring Velocity vs. Time 0 1 2 3 4 5 6 7 8 9 10 11 12 t(s) But what is the maximum velocity? Let's look at the data of position versus time for the mass in our example. t x 0.000 0.200 0.050 0.199 0.100 0.198 0.150 0.194 0.200 0.190 0.250 0.185 0.300 0.178 0.350 0.171 0.400 0.162 0.450 0.152 0.500 0.141 0.550 0.130 0.600 0.118 0.650 0.104 0.700 0.091 0.750 0.077 0.800 0.062 0.850 0.047 0.900 0.031 0.950 0.016 1.000 0.000 1.050-0.016 1.100-0.031 1.150-0.047 1.200-0.062 Look at the at the point where the position is x = 0.0m. Let us calculate the velocity.
v t = lim t 0 x t = lim t 0 x i 1 x i 1 t i 1 t i 1 = 0.016 m 0.016 m =.314 m 1.150 s 0.950 s s The magnitude of the maximum velocity is 0.314 m/s. Notice that: A = 0.2m 1.57 s 1 =0.314 m s So v max = A. Therefore we can model the velocity of the mass attached to the spring can be modeled as: v t = v max sin t = A sin t We can do the same analysis for the acceleration of the mass. Recall that the acceleration is the time rate change of the velocity. a t = lim t 0 v t = lim t 0 v i 1 v i 1 t i 1 t i 1 Now, let's look at a plot of acceleration (a) versus time (t). a(m/s2) 0.500 0.400 0.300 0.200 0.100 0.000-0.100-0.200-0.300-0.400 ass on a Spring Acceleration vs. Time -0.500 0.000 2.000 4.000 6.000 8.000 10.000 12.000 From the plot we can see that the acceleration as a function time can be modeled as a negative cosine function. a t = a max cos t t(s)
As it turns out the maximum acceleration is equal to: a max = A 2 So the acceleration as a function of time becomes: a t = a max cos t = A 2 cos t In summary, the motion of a mass attached to a spring on a frictionless table can be described by the following general equations. x t = A cos w t v t = v max sin w t = A sin w t a t = a max cos w t = A 2 cos w t For our example, the specific equations of motion, specific to our example that is, become: x t = 0.2 m cos 1.57 s 1 t v t = 0.314 m s sin 1.57 s 1 t a t = 0.493 m s 2 cos 1.57 s 1 t What else can we find about the motion of a mass on a spring? Consider the forces on the mass. F N F S =-kx W=mg Consider the forces perpendicular to the table, it is obvious that the normal force counteracts the weight, that is the magnitude of the normal force is equal to the magnitude of the weight. Now let's consider the forces parallel to the table.
F x = ma kx = ma k A cos t = m A 2 cos t k A cos t = m A 2 cos t k = m 2 = k m So the angular frequency is a function of the spring constant (k) and the mass (m). To find the period (T): = 2 T k 2 = m T T = 2 m k Exercise Two: Consider a 200 kg mass attached to a spring with a spring constant of 100 N/m. The mass and the spring sit horizontally on a frictionless table. The mass is moved to the position x = 0.4 m and released. The mass oscillates back and fourth. 1. Calculate the angular frequency and the period of the motion. 2. Plot the position of the mass versus time for two periods. 3. Use the method described above to calculate the velocity. Plot the velocity versus time. 4. Use the method described above to calculate the acceleration. Plot the acceleration versus time. 5. Calculate the maximum velocity. Find the percent error between the calculated value and the value you plotted. 6. Calculate the maximum acceleration. Find the percent error between the calculated value and the value you plotted. 7. Write the specific equations for the position, the velocity, and the acceleration as a function of time. ** See the Appendix A if you need help with spread sheeting mechanics. *** Hand in a copy of each plot with a cover sheet with your name and the answers to any questions asked. Staple them together.
Part Three ass on a Spring ( Another Look ) This section is for those of you who understand differential calculus. You'll see that the calculations are much cleaner and can be solved in closed form. Let's start with our model to describe the position of a mass attached to a spring. F s x=-a x=0.0 m x=+a x t = A cos t Now consider the velocity. The velocity is the time derivative of the position. v t = lim t 0 x t = dx dt v t = d dt [ A cos t ] = A sin t = v max sin t v max = A Now consider the acceleration. The acceleration is the time derivative of the velocity. a t = lim t 0 v t = dv dt a t = d dt [ A sin t ] = A 2 cos t = a max cos t a max = A 2
F x = ma kx = ma k A cos t = m A 2 cos t k A cos t = m A 2 cos t k = m 2 = k m