2.3 Oscillation. The harmonic oscillator equation is the differential equation. d 2 y dt 2 r y (r > 0). Its solutions have the form

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1 2. 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. There is a simple differential equation that leads to this behavior. The Harmonic Oscillator Equation The harmonic oscillator equation is the differential equation Its solutions have the form dt 2 r y (r > 0). y C cos(ωt) + D sin(ωt), where C and D are constants, and ω r. This is the Cartesian form of the solution. There is also a polar form that we will describe in a short while. As we will see shortly, the formula y C cos(ωt) + D sin(ωt) actually describes simple sinusoidal oscillation, also known as harmonic oscillation. The constant ω r is called the angular frequency of the oscillation. The square root comes from taking the second derivative; for if y C cos(ωt) + D sin(ωt) The angular frequency ω has units of radians per unit time, e.g. rad/sec. then taking the second derivative gives dt 2 and thus y satisfies the equation ω 2 C cos(ωt) ω 2 D sin(ωt) dt 2 ω 2 y. We conclude that r ω 2, and hence ω r. It would also work to let ω r, but this actually yields the same solutions as ω r, so it is simpler to use a positive value for ω. EXAMPLE 1 Solve the following initial value problem. We know that dt 2 25y, y(0), y (0) 10. y(t) C cos(5t) + D sin(5t) for some constants C and D, which means that y (t) 5C sin(5t) + 5D cos(5t). Plugging y(0) into the first equation and y (0) 10 into the second equation yields C Remember that cos(0) 1 and sin(0) 0. and D 2, so the solution is y(t) cos(5t) + 2 sin(5t).

2 2 OSCILLATION EXAMPLE 2 Solve the following boundary value problem. We know that ( ) π dt 2 y, y(0) 1, y 2 8 y(t) C cos(2t) + D sin(2t) for some constants C and D. The two boundary values yield the equations 1 C and ( ) ( ) π π 2 C cos + D sin. Plugging C 1 and cos(π/) sin(π/) 2 / 2 into the second equation and solving for D gives D 5, and hence y(t) cos(2t) + 5 sin(2t). Cartesian and Polar Forms Our general solution y C cos(ωt) + D sin(ωt) to the harmonic oscillator equation is called the Cartesian form of the solution. For many applications, it is more convenient to write the solution in the following polar form. Polar Form for the Solution The solutions to the harmonic oscillator equation dt 2 r y (r > 0). The constant A is called the amplitude, and φ is the phase angle. can also be written as where A and φ are constants. y A cos(ωt + φ) a Figure 1: The graph of a sinusoidal function. Here we used the identity cos(x + y) cos x cos y sin x sin y. Functions of the form y A cos(ωt + φ) are sometimes called sinusoidal functions. The graph of such a function is a simple sine wave, as shown in Figure 1. The solutions to the harmonic oscillator equation are precisely the sinusoidal functions, and any variable y that obeys the harmonic oscillator equation undergoes sinusoidal oscillation. It is not obvious that the Cartesian and polar forms of a sinusoidal function are equivalent. To see this, consider a sinusoidal function in polar form: y A cos(ωt + φ) Using the sum of angle formula for cosine, we can expand the right side to get y A cos(φ) cos(ωt) A sin(φ) sin(ωt). This has the form y C cos(ωt) + D sin(ωt), where

3 OSCILLATION C A cos(φ) and D A sin(φ) Conversely, for any values of C and D it is always possible to find values for A and φ that satisfy the above equations. In particular, A C 2 + D 2, cos(φ) C A, and sin(φ) D A The algebra here is similar to the conversion between rectangular and polar coordinates, where (x, y) (C, D) and (r, θ) (A, φ). These formulas let us convert sinusoidal functions between Cartesian and polar forms. EXAMPLE Find the Cartesian form of the sinusoidal function y(t) cos ( t + π ). We have A and φ π/, so ( ) ( ) π π C cos 2 and D sin 2 and therefore y(t) 2 cos(t) 2 sin(t). EXAMPLE Find the polar form of the sinusoidal function y(t) 6 cos(t) + 6 sin(t). We have C 6 and D 6, so A C 2 + D Then cos(φ) and sin(φ) , and hence φ π/. We conclude that y(t) 6 ( 2 cos t π ). Here we were able to just recognize the angle φ from its cosine and sine, but in general we can find φ using inverse trigonometric functions. Properties of Oscillation The two most important properties of any oscillation are its amplitude and its period. The amplitude of an oscillation is simply its maximium value A, as shown in Figure 2. For a sinusoidal oscillation, this is the coefficient of the cosine when the function is expressed in polar form. The period of an oscillation is the amount of time T that it takes for the oscillation to go through one complete cycle, as shown in Figure. For a sinusoidal oscillation, the period is given by the formula a Figure 2: The amplitude of an oscillation. T 2π ω where ω is the angular frequency. Closely related to the period of an oscillation is its frequency, which is defined by the formula a Figure : One period of an oscillation.

4 OSCILLATION f 1 T The SI unit for frequency is the hertz (Hz), where 1 Hz 1/sec. The frequency is measured in units of inverse time (i.e. number per unit time), and can be interpreted as the rate at which oscillations occur. For example, a sinusoidal function with a frequency of /sec undergoes three full oscillations each second, while a sinusoidal function with a frequency of 0.5/sec undergoes half of an oscillation each second, or one full oscillation every two seconds. Note that the frequency is not the same as the angular frequency. These are related by the formula ω 2π f Thus the angular frequency also measures the rate of oscillation, but it is much less natural than either the period or the frequency. Indeed, angular frequency is only really important because it appears in the formulas for sinusoidal oscillation. EXAMPLE 5 A harmonic oscillator satisfies the differential equation What is the period of oscillation? dt 2 0. y. The angular frequency is ω , so T 2π ω Finally, every sinusoidal oscillation has a phase angle φ, which describes the state of the oscillation when t 0. A phase angle of φ 0 corresponds to an oscillation that starts at its maximum value at t 0. Phase angles less than 0 correspond to starting earlier in the cycle, and phase angles greater than 0 correspond to starting later in the cycle, as shown in Figure. d Figure : Sinusoidal oscillations with four different phase angles φ.

5 OSCILLATION 5 A Closer Look Solving the Harmonic Oscillator Equation For ω > 0, we have stated without proof that the solutions to the equation dt 2 ω2 y are the functions y C cos(ωt) + D sin(ωt), where C and D can be any constants. It is easy to see that these functions are indeed solutions, but how can we be sure that every solution has this form? We can prove this as follows. Suppose that y(t) is a solution to the above equation, and define functions C(t) and D(t) by the formulas C(t) y(t) cos(ωt) y (t) ω sin(ωt), D(t) y(t) sin(ωt) + y (t) ω cos(ωt). Observe that y(t) C(t) cos(ωt) + D(t) sin(ωt). We wish to show that C(t) and D(t) are constant functions. To prove this, we take the derivative of each using the product rule. The derivative of C(t) is These formulas for C(t) and D(t) were obtained by solving the equations y C cos(ωt) + D sin(ωt) y ωc sin(ωt) + ωd cos(ωt) for C and D. C (t) y (t) cos(ωt) ωy(t) sin(ωt) y (t) ω The first and last terms cancel, leaving C (t) ωy(t) sin(ωt) y (t) ω sin(ωt) y (t) cos(ωt). sin(ωt). Substituting in y (t) ω 2 y(t) causes the two remaining terms to cancel, giving us C (t) 0. Thus C(t) is a constant function. A similar computation shows that D(t) is a constant function, and therefore y(t) has the desired form. EXERCISES 1. Solve the following initial value problem: dt 2 5y, y(0), y (0) Solve the following boundary value problem: dt 2 y, y(0), y ( 2π ) 1.. Solve the following boundary value problem: ( dt 2 y, y π ), 6. Express the sinusoidal function y(t) 10 cos ( ) π y ( 6t + π ) in Cartesian form.

6 6 OSCILLATION 5. Express each of the following sinusoidal functions in polar form. (a) y(t) cos(t) + sin(t) (b) y(t) sin(5t) (c) y(t) 2 cos t + 6 sin t 6. Find the amplitude of the following sinusoidal function: y(t) 12 cos(t) 5 sin(t). 7. Find the phase angle of the following sinusoidal function: Express your answer in degrees. y(t) cos(2t) + sin(2t). 8. Find the period of the following sinusoidal function: y(t) 7.2 cos(5.t + 1.2) 9. A sinusoidal function y(t) satisfies the differential equation dt y. What is the frequency of the oscillation? 10. A sinusoidal function y(t) has phase angle π/. Given that y(0) 5, what is the amplitude?

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