Physics 121H Fall 2015 Homework #14 16-November-2015 Due Date : 23-November-2015 Reading : Chapter 15 Note: Problems 7 & 8 are tutorials dealing with damped and driven oscillations, respectively. It may be useful to work the tutorials before the other problems. 1. A 1000 kg car carrying four 32 kg people travels over a "washboard" dirt road with corrugations 4.0 m apart. The car bounces with maximum amplitude when its speed is 16 km/h. When the car stops, and the people get out, by how much does the car body rise on its suspension? 2. Demonstrate that for a driven, damped oscillator the maximum position amplitude occurs at a frequency that is smaller than the natural frequency of the undamped undriven oscillator and that the maximum velocity amplitude occurs at the natural frequency. Recall that the position for a driven damped oscillator is given by. With the position amplitude,. ω d = driving frequency, F m = amplitude of driving force, b = damping constant and ω = natural frequency of undamped, undriven oscillator. 3. A damped oscillator is formed by attaching a mass with m = 1.5 kg to one end of a spring with spring constant k = 8 N/m. The other end of the spring is anchored and the mass can slide on a horizontal surface The damping force is given by bv with b = 230 g/s. At t=0, the mass is displaced so that the spring is compressed by 12 cm from its unstretched length and released from rest. (a) Find the time required for the amplitude of the resulting oscillations to decay to 1/3 of its initial value. (b) How many oscillations are made by the mass during this time? (c) Find the value of b so that the oscillator is critically damped. (d) At t=0, this critically damped oscillator is displaced so that the spring is stretched a distance of 12 cm beyond its unstretched length, find the time required for mass to reach the position for which the spring is streched by only 4 cm. 4. Show that the the time rate of change of the total mechanical energy of a damped, undriven oscillator is given by de/dt = -bv 2 where b is the damping constant. Assume that the oscillator is a mass m attached to a spring with constant k. Hint: Differentiate the total mechanical energy of the oscillator and use Newton s second law.
5. The center of oscillation of a physical pendulum has this interesting property: If an impulse (assumed horizontal and in the plane of oscillation) acts at the center of oscillation, no oscillations are felt at the point of support. Baseball players (and players of many other sports) know that unless the ball hits the bat at this point (called the "sweet spot" by athletes), the oscillations due to the impact will sting their hands. To prove this property, let the uniform rod shown in the figure simulate a baseball bat. Suppose that a horizontal force F (due to impact with the ball) acts toward the right at P, the center of oscillation. The batter is assumed to hold the bat at O, the pivot point of the stick. a) What acceleration does the point O undergo as a result of F? (b) What angular acceleration is produced by F about the center of mass of the stick? (c) As a result of the angular acceleration in (b), what linear acceleration does point O undergo? (d) Considering the magnitudes and directions of the accelerations in (a) and (c), convince yourself that P is indeed the "sweet spot." 6. The suspension system of a 2000 kg automobile "sags" 10 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 50% each cycle. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 500 kg. 7. This problem is a tutorial about damped oscillations. As you work through the tutorial, you will be asked to demonstrate several important points. You're supposed to turn in these demonstrations as part of this week's HW. The demonstrations are numbered '7.X' in the following. The tutorial refers to the figure, which shows a mass that is free to slide on a horizontal surface. The mass is attached to a spring and the other end of the spring is attached to a fixed wall. We will measure the position of the mass using 'x' so that x=0 when the spring is at its relaxed length and +x is in the direction of spring extension. In addition to the spring force from last week, the mass experiences a damping force that is proportional to and in the opposite direction of the velocity,. b is called the damping constant. With this damping force, Newton's second law for the mass reads (remember that the weight and normal force cancel out). Equating the x-components of the vectors and rearranging gives ma + bv + kx = 0. Recalling that the velocity and acceleration are both time derivatives of the position, we find the differential equation to be solved. (1) Just like last week, this is a 2 nd order, linear ordinary differential equation with constant coefficients. So, just like last week, it must have a solution of the form. (2)
Again, like last week, our job is to find the values for the constants A and λ so that eqn. (2) solves eqn. (1). 7.1 By substituting eqn. (2) into eqn. (1), find the characteristic equation that we need to solve for λ: (mλ 2 + bλ + k)x = 0. Since x=0 can only be true if the mass never moves, the quantity in brackets must be zero. 7.2 In this case, show that the allowed values for λ are (3) where ω o 2 = k/m is the natural frequency of the undamped oscillator (it's really the natural angular frequency, but we will be lazy). Again, like last week, since the differential equation is 2 nd order, there must be 2 roots for the characteristic eqn. Note that there are 3 distinct cases, all of which have physical significance, as we shall see. (i) (underdamped) (ii) (critically damped) (iii) (over damped). We will consider each case in order, starting with underdamped oscillators. In this case, the quantity under the radical in eqn. (3) is negative. It is convenient to define, as we will see this ω is the angular frequency of the underdamped oscillator. With this definition for ω, eqn. (3) becomes =. 7.3 Substitute these values for λ ± into eqn. (2) to get. (4) 7.4 Follow the same steps as last week (9.5 & 9.6) to turn the quantity in brackets in eqn. (4) into A cos(ωt + φ) to show the solution for the undamped oscillator is
x(t) = A e -bt/2m cos(ωt + φ), (underdamped). (5) Note that the (angular) frequency of the damped oscillator, ω, is less than that of the undamped oscillator. Note also that the exponential function in front of the cosine ensures that motion never repeats itself. So, it doesn't make any sense to refer to the period of damped oscillations. Rather, we will refer to an analogous quantity, the 'cycle time', 'T' = 2π/ω. The cycle time is the time between successive maxima, successive minima or every other zero-crossing of the x(t) curve. Note that the amplitude of the oscillations steadily decreases as time goes on according to A(t) = A e -bt/2m. This function, A(t) is known as the envelop function of the underdamped oscillator since it 'wraps' the cosine function in an exponentially decreasing envelop. Since the kinetic energy of the oscillator is zero when the x(t) curve is at either a maximum or minimum, the total mechanical energy of the oscillator is all elastic potential energy at these times. 7.5 Show that it is OK to estimate the total mechanical energy of an underdamped oscillator using E(t) = ½ k A 2 e -bt/m. (6) This approximation for E(t) is best for lightly damped (i.e., b is small) oscillators. See the course web page for some examples of underdamped oscillation x(t) curves with their envelop functions shown and for some E(t) curves that demonstrate the validity of eqn. (6) and why it is a better approximation for lightly damped oscillators. For all of these curves, the oscillator is released from rest at x = 1. Next, we will consider critically damped oscillators. In this case, the quantity under the radical in eqn. (3) is zero. This means we have only one root for λ, but we need two. Fortunately, the mathematicians come to our rescue and tell us that it is OK to form the solution like this (A and B are both constants) 7.6 By direct substitution, show that eqn. (7) solves eqn. (1). (critical damping). (7) The critically damped solution is the solution that reaches equilibrium (x=0) the fastest without overshooting if the oscillator is released from rest. This solution is plotted and discussed on the course web page. Finally, we will consider overdamped oscillators. In this case, the quantity under the radical sign in eqn. (3) is positive and the solution is (A and B are both constants) (over damped). (8) When plotted, the overdamped solutions appear very similar to the critically damped solution, they just take longer to reach equilibrium. There are plots of several solutions
on the course web page that compare under, critical and overdamping so you can see how they look. 8. This is a tutorial about driven, damped oscillations and resonance. We will think about the same oscillator as the previous tutorial. As you work through the tutorial, you will be asked to demonstrate several important points. You're supposed to turn in these demonstrations as part of this week's HW. The demonstrations are numbered '8.X' in the following. Instead of displacing the oscillator from equilibrium and releasing it from rest, we will 'drive' it by applying this periodic force to the mass. (1) Including this force in Newton's 2 nd law gives the amplitude of the driving force and ω d is the driving frequency.. F o is 8.1 Rearrange Newton's 2 nd law, equate the x-components of the vectors and recall that v and a are time derivatives of the position to get. (2) This differential equation is inhomogeneous. i.e., the right-hand-side (RHS) 0. These types of equations are more challenging to solve, so we will cheat. We can cheat since we are only interested in the 'steady state' solution. Steady state refers to the oscillations a long time after the driving force was 'switched on' when the oscillator's response is no longer changing. As we will see in lab 8, this means that the oscillation is at the driving frequency, ω d, has constant amplitude and constant phase difference to the driving force. We will find that both the amplitude and phase angle depend on the driving frequency. So, here is our strategy. We will guess that the position of the mass in steady state is. This probably seems a little goofy, since according to Euler's equation, this guess for x will have an imaginary part. But that's OK, since we will just 'throw away' the imaginary part and keep the real part. This means that we need to be very careful about keeping track of the real and imaginary parts of the complex numbers as we work. Here's the 'trick'. Every complex number, z = a + ib, has both real and imaginary parts. Both the real and imaginary parts are real numbers. We write Re(z) = a and Im(z) = b. Re() and Im() mean to take the real or imaginary part of the complex number inside the (). 8.2 Assming that A is a real number, use Euler's equation to show that Re(x) = A cos(ω d t) and Im(x) = A sin(ω d t).
By the same reasoning, if F o is a real number (it must be) we can write the RHS of eqn. (2) as Re( ). Now we will get started by replacing the RHS of eqn.(2) with Re( ). Next, we'll insert our guess for x(t) into the LHS of eqn.(2) 8.3 Show that and. Now we will use these results to rewrite eqn. (2) as. (3) From here to the final solution, we just need to be very careful about separating the real and imaginary parts of all the complex numbers. First note that the exponential functions cancel out on both sides of eqn. (3). This makes it really simple to solve for the driving frequency dependent amplitude. 8.4 Remember that the natural frequency of the undamped, undriven oscillator is ω o = (k/m) ½. Use this fact to show that after cancelling the exponential functions in eqn. (3), the amplitude can be written as. (4) This is progress, but we need to separate this amplitude into its real and imaginary parts. Here's how. We will just multiply by one. We will use the following fact. 8.5 Show that if z is some complex number z = a + ib (remember that a and b are real numbers) and if we define the complex conjugate of z as z* = a ib. Then by multiplying any complex number by its complex conjugate, we get the square of its magnuitude: z z* = a 2 + b 2. Use this fact to show (this is still part of 8.5). (5) With eqn. (5) we can move the 'i' in the denominator of eqn. (4) up to the numerator, which makes it oodles easier to keep track of the real and imaginary parts. 8.6 Use eqn. (5) to show that eqn. (4) can be rewritten as. (6) Now, all we have to do is to multiply the expression for A(ω d ) in eqn. (6) by to get our solution for x(t). We're not done, though since we need to separate x(t) into its real
and imaginary parts. This is a little painful, but not too bad. 8.7 Show that = Re[A(ω d )]cos(ω d t) Im[A(ω d )]sin(ω d t). Now, use this result to show that the required real part of x(t) is. (7) Now, we are close. Focus on the quantity in [] in eqn. (7). 8.8 Set and. Use a trig identity to rewrite the quantity in [] in eqn. (7) as. Refer to the diagram at the right to show that (8) and. (9) Now we just have to put it all together. 8.9 Substitute eqn. (8) and eqn. (9) into eqn. (7) to get the final answer (10) with. (11) and φ(ω d ) given by eqn. (9). 8.10 Eqn. (10) describes the position of the oscillator at any time in steady state. What is the expression for the velocity? You should get (some constant) multiplied by (some trigonometric function). The constant out in front of the trigonometric function is called the 'velocity amplitude'. Some plots displaying the driving frequency dependence of the amplitude and phase angle are on the course web page.