2.1 Exponential Growth

Transcription

1 2.1 Exponential Growth A mathematical model is a description of a real-world system using mathematical language and ideas. Differential equations are fundamental to modern science and engineering. Many of the known laws of physics and chemistry are expressed as differential equations, and differential equation models are used extensively in biology, economics, and other social sciences. The most common use of differential equations in science is to model dynamical systems, i.e. systems that change in time according to some fixed rule. For such a system, the variable under consideration (such as mass M, population P, or electric charge Q) is a function of time t. Thus, a typical differential equation might be written dm t 3 M 2 instead of the more familiar y x 3 y 2. The Exponential Growth Equation Perhaps the most common differential equation in science is the equation for exponential growth. It can be used to model a variety of phenomena in the natural and social sciences, including unrestricted population growth, nuclear chain reactions, continuously compounded interest, and macroeconomic growth. The Exponential Growth Equation The exponential growth equation is the differential equation dy k y, where k is a positive constant. Its solutions have the form where y 0 y(0) is the initial value of y. y y 0 e kt, The formula y y 0 e kt represents exponential growth, as shown in Figure 1. The constant k is called the growth constant, and has units of inverse time (e.g. number per second). a Figure 1: Exponential growth. EXAMPLE 1 Solve the following initial value problem. dy 3y, y(0) 4. According to the formulas above, the solution is y(t) 4e 3t.

2 EXPONENTIAL GROWTH 2 EXAMPLE 2 Solve the following initial value problem. dy 3y, y(2) 8. This time the initial value is at t 2 instead of t 0, so we have some work to do. We know that y has the form y(t) y 0 e 3t for some constant y 0. Plugging in y(2) 8 gives 8 y 0 e 6, so y 0 8e 6. Then y(t) ( 8e 6) e 3t 8e 3t 6. a Figure 2: Exponential population growth. Population Growth Exponential growth is often used to model the growth of populations of organisms in a resource-rich environment. Here resource-rich means that there is plenty of food and other resources necessary for the population to grow indefinitely. The relative growth rate of a population P(t) is the quantity r 1 P dp. This represents the rate of growth of P relative to the size of the population. For example, if a bacteria culture has a relative growth rate of r 3/hour, it means that each individual bacterium has an average of 3 offspring per hour (not counting grandchildren). Assuming a population P mantains a constant relative growth rate, its growth will satisfy the differential equation dp rp. Such a population grows exponentially with time: P P 0 e rt where P 0 is the population at time t 0 (see Figure 2). EXAMPLE 3 During a biology experiment, a certain culture of cells has a constant relative growth rate of 0.04/minute. If there are 5,000 cells at the beginning of the experiment, how large will the culture be one hour later? The population P(t) is given by the equation P(t) 5,000e 0.04t. We are measuring t in minutes, so one hour later corresponds to t 60. Since P(60) 5,000e 0.04(60) 55,115.9, the population after one hour will be approximately 55,000 cells.

3 EXPONENTIAL GROWTH 3 EXAMPLE 4 A population of bacteria initially consists of 20,000 cells. Twenty minutes later, the population has grown to 50,000 cells. How quickly is the population increasing at that time? equation Assuming constant relative growth rate, the population P(t) is given by the P(t) 20,000e rt. (1) for some constant r. Plugging in P(20) 50,000 gives the equation To solve the equation 20,000e 20r 50,000 for r, we first divide by 20,000 to get e 20r 2.5, and then we take the natural logarithm of both sides, which yields 20r ln(2.5). We can now divide through by 20 to obtain the value of r. 50,000 20,000e r(20). and solving for r yields ln 2.5 r To find the rate of increase at t 20 min, we must compute P (20). Taking the derivative of equation (1) gives P (t) 20,000r e rt so P (20) 20,000( ) e ( )(20) Thus the population is increasing at a rate of roughly 2,300 cells/min. Describing the Growth Rate For an exponential growing quantity y y 0 e kt, there are a few alternatives to using the growth constant k to describe the rate of exponential growth. The simplest is to use the time constant τ, which is the reciprocal of the growth constant: τ 1 k. Using the time constant, the formula for exponential growth can be written y y 0 e t/τ. The time constant τ has units of time, and represents the amount of time that it takes for the variable y to be multiplied by a factor of e. For this reason, τ is also known as the e-folding time. Another way to describe the rate of exponential growth is to use the doubling time T d, which is the amount of time that it takes for y to double in size. The doubling time satisfies the equation e kt d 2 and hence T d ln 2 (ln 2)τ. k Since ln , the doubling time is approximately 69% of the e-folding time. EXAMPLE 5 A small tumor is growing exponentially with a doubling time of 15 days. If the tumor currently has 1.3 million cells, how quickly is the number of cells increasing? formula Since the tumor is growing exponentially, the number of cells N is given by the N(t) N 0 e kt,

4 EXPONENTIAL GROWTH 4 where N million cells. Since the doubling time is 15 days, we know that e 15k 2. Taking the natural logarithm of both sides and then dividing through by 15 gives k /day. To find N (0), we take the derivative of the formula for N to get N (t) N 0 k e kt. Plugging in t 0 gives N (0) N 0 k (1.3 million cells)( /day) 60 thousand cells/day. EXERCISES 1. Solve the following initial value problem: dy 3y, y(ln 2) At the beginning of an experiment, a culture of E. coli contains 15,000 cells. Two hours later, the population has grown to a size of 80,000 cells. Assuming exponential growth, what is the time constant τ? 3. During a nuclear chain reaction, the number N of free neutrons in a sample of fissile material obeys the differential equation dn kn, where k a constant known as the neutron multiplication factor. (a) Suppose the sample initially contains 100 free neutrons. If the neutron multiplication factor is 7.0/µs, how many free neutrons will there be 2.0 µs later? (b) How quickly will the number of free neutrons be increasing at that time? 4. The population of the United States is currently 320 million, and is increasing at a rate of 2.4 million/year. Assuming exponential growth, what is the doubling time of the population? 5. A cell culture is growing exponentially with a doubling time of 3.00 hours. If there are 11,500 cells initially, how long will it take for the cell culture to grow to 30,000 cells?

5 2.2 Exponential Decay When the growth constant k is negative, the function y y 0 e kt does not actually grow. Instead, the value of y approaches zero as t. This is known as exponential decay. When discussing exponential decay, it is common in applications to write k r, where r is a positive constant known as the decay constant. In this case, the differential equation takes the following form. The Exponential Decay Equation The exponential decay equation is the differential equation This is just the exponential growth equation with r substituted in for k. dy r y where r is a positive constant. Its solutions have the form where y 0 y(0) is the initial value of y. y y 0 e rt a Figure 1: Exponential decay. See the EPA website for a list of radioisotopes commonly used in industry. a Figure 2: Exponential decay of a radioactive sample. Figure 1 shows the graph of a typical exponential decay. As with exponential growth, there are two alternatives to using the decay constant r when describing exponential decay: 1. The time constant (or e-folding time) of y is the quantity τ 1/r, and represents the amount of time that it takes for the value of y to be divided by e. 2. The half-life of y is the amount of time that it takes for the value of y to be cut in half. As we shall see, exponential decay can be used to model such diverse phenomena as radioactive decay, electric circuits, and chemical reactions. Radioactive Decay A substance whose atoms are inherently unstable is called radioactive. For such a substance, a certain fixed proportion of the atoms decay during each time interval. If N is the number of atoms in a sample of the substance, then N will satisfy the differential equation dn rn. Here r is the decay constant, which represents the rate at which individual atoms tend to decay. For example, if r 0.003/hour, it means that about 0.3% of the atoms will decay each hour. The number of atoms in a radioactive sample decays exponentially with time: N N 0 e rt, where N 0 is the number of atoms int the sample at time t 0 (see Figure 2). Equivalently, the total mass M of atoms in a sample will decay exponentially: M M 0 e rt.

6 EXPONENTIAL DECAY 2 EXAMPLE 1 Cesium-137 has a half-life of approximately years. If a mole sample of 137 Cs is left in a storage closet, how much 137 Cs will be left after four years? The amount N(t) of 137 Cs will obey an equation of the form N(t) 0.30e kt, where k is a constant. Since the half-life is years, we know that e k(30.17) 1 2. Solving for k gives k /year. Then N(4) 0.300e ( )(4) , so there should be moles left after four years. EXAMPLE 2 A radiochemist prepares a cobalt sample containing moles of 58 Co. According to readings from a Geiger counter, the atoms of 58 Co in the sample appear to be decaying at a rate of moles/min. Based on this information, what is the half-life of 58 Co? The amount N(t) of 137 Cs will obey an equation of the form N(t) 0.100e kt, N (0) is negative since N is decreasing. where k is a constant. Then N (t) 0.100k e kt. We are given that N (0) moles/min, which gives us the equation k. Solving for k gives k /min. The half-life is the value of t for which e kt 1 2. Plugging in k and solving for t yields a half life of 102,084 minutes, which is about 70.9 days. resistor capacitor RC Circuits Figure 3 shows a simple kind of electric circuit known as an RC circuit. This circuit has two components: A resistor is any circuit component such as a light bulb that resists the flow of electric charge. Applying voltage to a resistor will force current through it, with the amount of current given by Ohm s law I V R. (1) a Figure 3: An RC circuit. Here V is the applied voltage, I is the resulting current, and R is a constant called the resistance of the resistor.

7 EXPONENTIAL DECAY 3 A capacitor is a circuit component that stores electric charge. A charged capacitor can supply voltage to a circuit, with the amount of voltage given by the equation V Q C. (2) Here Q is the charge stored in the capacitor and C is a constant called the capacitance of the capacitor. All of the circuit-related quantities discussed here have their own SI units: Electric charge is measured in coulombs (C). Voltage is measured in volts (V). Electric current is measured in amps (A), where 1 A 1 C/sec. Resistance is measured in ohms (Ω), where 1 Ω 1 V/A. Capacitance is measure in farads (F), where 1 F 1 C/V. In an RC circuit, the voltage produced by a capacitor is applied directly across a resistor. Substituting equation (2) into equation (1) yields a formula for the resulting current: I Q RC. (3) This current represents the flow of charge out of the capacitor, with dq I. Substituting equation (3) into this equation yields the differential equation dq Q RC. Thus the charge Q will decay exponentially, with decay constant Equivalently, the time constant for the RC circuit is τ RC. r 1 RC In the case where the resistor is a light bulb, the result is that the bulb lights up at first, but becomes dimmer and dimmer over time as the capacitor discharges. EXAMPLE 3 A 0.25 F capacitor holding a charge of 2.0 C is attached to a 1.6 Ω resistor. How long will it take for the capacitor to expend 1.5 C of its initial charge? The charge on the capacitor will decay exponentially according to the formula Q Q 0 e rt, where Q C and Since seconds are the SI unit of time, the units for r must be number per second. r 1 RC 1 (1.6 Ω)(0.25 F) 2.5/sec. If the capacitor expends 1.5 C of its charge, it will have 0.5 C left. Plugging this into the formula for Q gives e (2.5)t, and solving for t yields t Thus the capacitor will expend 1.5 C of charge in approximately 0.55 seconds.

8 EXPONENTIAL DECAY 4 EXERCISES 1. A sample of an unknown radioactive isotope initially weighs 5.00 g. One year later the mass has decreased to 4.27 g. (a) How quickly is the mass of the isotope decreasing at that time? (b) What is the half life of the isotope? 2. During a certain chemical reaction, the concentration [C 4 H 9 Cl] of butyl chloride obeys the rate equation d[c 4 H 9 Cl] r [C 4 H 9 Cl], where r /sec is the rate constant for the reaction. How long will it take for this reaction to consume 90% of the initial butyl chloride? 3. A capacitor with a capacitance of 5.0 F holds an initial charge of 350 C. The capacitor is attached to a resistor with a resistance of 9.0 Ω. (a) How quickly will the charge held by the capacitor initially decrease? (b) How quickly will the charge be decreasing after 20 seconds? 4. An LR circuit consists of a resistor attached to an electrical component called an inductor, which supplies voltage to the resistor according to the formula V L di. Here L is a constant called the inductance of the inductor. (a) Combine the equation above with Ohm s law to obtain a differential equation for the current I(t) that involves the constants L and R. (b) The current I(t) in an LR circuit decays exponentially. Find a formula for the decay constant in terms of L and R. The first of these equations is the molar form of the ideal gas law, while the second is the equation for hydrostatic pressure. 5. For a planetary atmosphere of ideal gas of uniform temperature T, the atmospheric pressure P(h) and density ρ(h) at a height h above the ground are related by the equations dp P ρrt and ρg, dh where R is the specific gas constant and g is the acceleration due to gravity. (a) Combine the given equations to obtain a single differential equation for P involving the constants g, R, and T. (b) The atmospheric pressure in such an atmosphere varies with height according to the formula P(h) P 0 e rh, where P 0 is the pressure at ground level. Find a formula for the decay constant r in terms of g, R, and T.

9 2.3 Harmonic Oscillation So far, we have used differential equations to describe functions that grow or decay over time. The next most common behavior for a function is to oscillate, meaning that it increases and decreases in a repeating pattern. The simplest kind of oscillation is harmonic oscillation. The square root in the formula for ω comes from taking the derivative twice. Specifically, if then and so y A cos(ωt + φ), y ωa sin(ωt + φ), y ω 2 A cos(ωt + φ), y ω 2 y. The Harmonic Oscillator Equation The harmonic oscillator equation is the differential equation d 2 y 2 r y, where r is a positive constant. Its solutions have the form y A cos(ωt + φ), where ω r, and A 0 and π < φ π are constants. a Figure 1: Harmonic oscillation. Figure 1 shows a typical harmonic oscillation. The constant A is the amplitude of the oscillation, and is equal to the maximum value of y. The constant φ is the phase constant, which is measured in radians, and the constant ω is the angular frequency, which is measured in radians per unit time. There are a few common alternatives to angular frequency for measuring the rate of oscillation: 1. The period T is the amount of time it takes for the oscillation to complete one full cycle. It is related to the angular frequency of a harmonic oscillation by the formula T 2π ω The SI unit for frequency is the hertz (Hz), which represents one cycle per second. 2. The frequency f is the number of cycles per unit time. The frequency is related to the period T and the angular frequency ω for a harmonic oscillation by the formulas f 1 T ω 2π EXAMPLE 1 A harmonic oscillator satisfies the differential equation What is the period of oscillation? d 2 y y. The angular frequency is ω , so T 2π ω

10 HARMONIC OSCILLATION 2 Cartesian Form There is a nice alternate form for harmonic oscillations that involves two terms added together. Harmonic Oscillations (Cartesian Form) The general solution to the differential equation can also be written as d 2 y 2 r y, y C 1 cos(ωt) + C 2 sin(ωt), where ω r and C 1 and C 2 are arbitrary constants. Given a harmonic oscillator y A cos(ωt + φ), Here we are using the formula cos(a + b) cos a cos b sin a sin b. we can convert it to Cartesian form by applying the angle sum formula for cosine: Thus y C 1 cos(ωt) + C 2 sin(ωt), where y A cos φ cos(ωt) A sin φ sin(ωt). C 1 A cos φ and C 2 A sin φ Conversely, given the values for C 1 and C 2, we can find the amplitude A and phase constant φ using the formulas A C C 2 2, cos φ C 1 A, sin φ C 2 A. The advantage of Cartesian form is that it vastly simplifies the task of solving initial value and boundary value problems. EXAMPLE 2 Solve the following initial value problem. d 2 y 2 25y, y(0) 3, y (0) 10. We know that y(t) C 1 cos(5t) + C 2 sin(5t) for some constants C 1 and C 2, which means that y (t) 5C 1 sin(2t) + 5C 2 cos(2t). In general, if y C 1 cos(ωt) + C 2 sin(ωt) then y(0) C 1 and y (0) ωc 2. Plugging y(0) 3 into the first equation and y (0) 10 into the second equation yields C 1 3 and C 2 2, so the solution is y(t) 3 cos(5t) + 2 sin(5t).

11 HARMONIC OSCILLATION 3 EXAMPLE 3 Solve the following boundary value problem. d 2 y ( ) π 2 4y, y(0) 1, y We know that y(t) C 1 cos(2t) + C 2 sin(2t) for some constants C 1 and C 2. The two boundary values yield the equations 1 C 1 and 3 ( ) ( ) π π 2 C 1 cos + C 2 sin. 4 4 Plugging C 1 1 into the second equation and solving for C 2 gives C 2 5, and hence y(t) cos(2t) + 5 sin(2t). EXERCISES 1. A harmonic oscillator y(t) has phase constant π/3. Given that y(0) 5, what is the amplitude? ( 2. Express the harmonic oscillator y(t) 10 cos 6t + π ) in Cartesian form Find the amplitude and phase constant for each of the following harmonic oscillators. (a) y(t) cos(3t) + sin(3t) (b) y(t) 3 cos(2t) + 4 sin(2t) (c) y(t) sin(5t) (d) y(t) 2 cos t + 6 sin t 4. Solve the following initial value problem: d 2 y 2 5y, y(0) 3, y (0) Solve the following boundary value problem: d 2 y 2 y, y(0) 4, y ( 2π 3 ) Solve the following boundary value problem: d 2 y ( 2 4y, y π ) 4, 6 ( ) π y 16. 6

12 2.4 Harmonic Oscillator Models In this section we give three important examples from physics of harmonic oscillator models. Such models are ubiquitous in physics, but are also used in chemistry, biology, and the social sciences to model oscillatory behavior. spring block frictionless table a Figure 1: A simple mass-spring system. We are neglecting friction in this computation. Adding a frictional force to the model would make this an example of damped harmonic oscillations. An object whose position oscillates harmonically is said to undergo harmonic motion. Mass-Spring Systems Consider the simple mass-spring system shown in Figure 1, which consists of a block with mass m attached to spring whose other end is fixed. According to Hooke s Law, the force that the spring exerts on the block is given by the equation F kx. Here k is the spring constant, and x is the horizontal position of the block, with x 0 being the rest position. Newton s second law (F ma) can be written as F m d2 x 2, where x(t) is the horizontal position of the block. Assuming the spring is the only horizontal force affecting the block, this gives us the differential equation kx m d2 x 2, which we can rewrite as d 2 ( ) x k 2 x. m This is a form of the harmonic oscillator equation, with angular frequency ω k m EXAMPLE 1 A 3.0 kg mass is attached to a spring with a spring constant of 4.0 kg/sec 2. The spring is stretched 0.80 m from its rest position and then the mass is released. What is the speed of the mass 1.0 sec later? The mass will undergo harmonic motion with an angular frequency of k 4.0 ω m rad/sec. 3.0 The position x(t) of the mass obeys a formula of the form x(t) C 1 cos(ωt) + C 2 sin(ωt), for some constants C 1 and C 2. Taking the derivative gives x (t) ωc 1 sin(ωt) + ωc 2 cos(ωt) We are given that x(0) 0.80 and x (0) 0, and plugging these in gives C and C 2 0, so the speed of the mass at t 1.0 sec is x (1.0) (1.1547)(0.80) sin(1.1547) 0.84 m/sec.

13 HARMONIC OSCILLATOR MODELS 2 inductor a Figure 2: An LC circuit. capacitor The SI unit of inductance is the henry (H), where 1 H 1 V sec/a. LC Circuits Figure 2 shows a simple kind of electric circuit known as an LC circuit. This circuit consists of a capacitor connected to a circuit component inductor, which is essentially just a coil of wire. Unlike a resistor, which always resists the flow of current, an inductor tends to oppose changes to the flow of electric current. That is, it s difficult to start pushing current through an inductor, but once the current gets going, it s difficult to make it stop. The voltage drop V across an inductor is given by the formula V L di Here L is a constant called the inductance of the inductor. Note that the inductor has positive voltage drop (like a resistor) when the current is increasing, but when the current is decreasing the voltage drop is negative, meaning that the inductor actually pulls current through it. Combining the formula above with the equation V Q/C for the capacitor yields L di Q C. As in an RC circuit, the electric current I is that same as the rate at which the capacitor is discharging, so I dq. Substituting this into the previous equation yields the differential equation L d2 Q 2 Q C. which we can rewrite as d 2 ( ) Q 1 2 Q. LC This equation describes simple harmonic oscillation with angular frequency ω 1 LC Thus the charge held in the capacitor oscillates according to the formula Q A cos(ωt + φ), The capacitor is like the spring in an LC circuit (with spring constant 1/C), while the inductor is the mass. where A is the amplitude of the oscillations (measured in coulombs, the SI unit of charge), and φ is the initial phase of the circuit. Roughly speaking, if we assume that the capacitor begins charged, then the capacitor begins by discharging through the inductor, slowly at first but picking up speed as the inductor lets more current through. Once the capacitor is fully discharged, the inductor continues pushing current through the circuit, which drains even more charge from the capacitor, leaving it with a negative total charge. The capacitor then reverses the flow of current to regain the lost charge, but the same thing happens again, with the inductor continuing to push current through in the reverse direction until the capacitor is back to its initial charged state. The cycle thus continues indefinitely.

14 HARMONIC OSCILLATOR MODELS 3 EXAMPLE 2 An LC circuit consists of a capacitor with a capacitance of F and an inductor with an inductance of 0.10 H. The capacitor starts with an initial charge of 0.12 C, and the initial current is zero. What is the magnitude of the current in the circuit 0.10 seconds later? The charge stored on the capacitor will oscillate harmonically with ω 1 LC 1 (0.016)(0.10) 25 rad/sec. Then for some constants C 1 and C 2, with derivative Q(t) C 1 cos(25t) + C 2 sin(25t) Q (t) 25C 1 sin(25t) + 25C 2 cos(25t). We are given that Q(0) 0.12 and Q (0) 0, and plugging these in gives C and C 2 0. Then Q (0.10) (25)(0.12) sin(2.5) Thus the current at time t 0.10 sec is approximately 1.8 A. pivot rod Pendulums A pendulum consists of a mass suspended from a rod that swings from a fixed pivot point, as shown in Figure 3. If θ(t) denotes the angle of the string from the vertical, then θ obeys the differential equation d 2 θ 2 g sin θ L, (1) a Figure 3: A pendulum. mass where g is the acceleration due to gravity (usually 9.8 m/sec 2 ), and L is the length of the string. Unfortunately, equation (1) is not an instance of the harmonic oscillator equation, because the right side involves sin θ instead of θ. This means that a pendulum is actually an anharmonic oscillator. For example, Figure 4 shows the noticeably anharmonic motion of a pendulum that starts from rest at θ(0) 0.99π. Though the motion of a pendulum is anharmonic, we can make a harmonic approximation for the motion in the case where θ isn t too large. This depends on the linear approximation sin θ θ, a Figure 4: Anharmonic motion of a pendulum with initial conditions θ(0) 0.99π and θ (0) 0. which is quite accurate when θ is close to zero, as shown in Figure 5. Indeed, as long as θ stays between 14 and 14, this approximation is accurate to within 1%. Replacing sin θ with θ in equation (1) gives us the approximate differential equation d 2 θ g ) 2 ( θ. L This equation describes approximate harmonic motion, with angular frequency a Figure 5: The graphs of y sin θ and y θ nearly coincide for θ close to zero. ω g L

15 HARMONIC OSCILLATOR MODELS 4 EXAMPLE 3 A swinging pendulum with a length of 2.0 m has an initial position of θ(0) 0.10 rad and an initial angular velocity of θ (0) 0.12 rad/sec. What will the position of the pendulum be 0.80 sec later? Then The pendulum will move approximately harmonically with angular velocity g 9.8 ω L rad/sec2. θ(t) C 1 cos(ωt) + C 2 sin(ωt) and θ (t) ωc 1 sin(ωt) + ωc 2 cos(ωt) for some constants C 1 and C 2. Plugging in the initial conditions gives C rad and C rad. Then θ(0.50) (0.10) cos( ) ( ) sin( ) , so the pendulum will be at an angle of approximately rad.

16 HARMONIC OSCILLATOR MODELS 5 EXERCISES 1. A pendulum with a length of 0.30 m starts from rest at an angle of 0.18 rad. How quickly will the angle of the pendulum be changing 0.20 sec later? 2. A 2.0 µf capacitor is connected to an inductor. If the resulting system oscillates with a frequency of 3.0 khz, what is the inductance of the inductor? 3. A 3.5 kg mass has been attached to a spring with a spring constant of 24 kg/sec 2. If the mass is oscillating with an amplitude of 1.6 m, what is the maximum speed of the mass during the oscillation? 4. A charged capacitor with a capacitance of 3.0 F is attached to an inductor with an inductance of 0.20 H. The initial current in the circuit is zero, but after 1.0 sec the current has increased to 8.0 A. What was the initial charge on the capacitor? 5. When two masses m 1 and m 2 are connected by a spring, the length L of the spring obeys the differential equation m 1 m 2 m 1 + m 2 d 2 L 2 kl, where k is the spring constant. (a) Suppose a 3.0 kg mass is attached to a 5.0 kg mass by a spring with a spring constant of 12 kg/sec 2. What is the period of the resulting oscillations? (b) In a carbon monoxide (CO) molecule, the bond between the carbon and oxygen atoms can be modeled as a spring with spring constant u/sec 2. Given that the carbon atom has a mass of 12.0 u and the oxygen atom has a mass of 16.0 u, at what frequency will the molecule tend to vibrate? 6. A pendulum with a length of 20.0 cm is started from rest at an angle of (a) What period does the harmonic approximation predict for this pendulum? Assume that the acceleration due to gravity is 9.81 m/sec 2. (b) The pendulum is measured to have an actual period of seconds. What was the percentage error in the period predicted by the harmonic approximation?