K. - Simple Harmonic Motion Chapter K Oscillatory Motion Blinn Collee - Physics 2425 - Terry Honan The Mass-Sprin System Interactive Fiure Consider a mass slidin without friction on a horizontal surface. The force of a sprin is iven by Hooke's law F = -k x. Applyin Newton's second law ives a second order ODE Definin the anular frequency by F net = m a -k x = m d2 x d t 2 k m, ives the second order ODE 0 = d t + 2 2 x. The Differential Equation for Simple Harmonic Motion We need to distinuish our standard example of simple harmonic motion, the mass-sprin system, from the eneral phenomenon. Generally simple harmonic motion occurs any time a mechanical system ives rise to a differential equation of the form We will see that this is quite a eneric expression. 0 = d t + 2 2 x.
2 Chapter K - Oscillatory Motion Comments on ODEs (Ordinary Differential Equations) A differential equation (DE) is some equation involvin a function and its derivatives. The differential equation is solved to find the function. The order of a DE is the hihest number of derivatives. If there is at most a second derivative it is a second order equation. If it is a function of one variable it is an ordinary differential equation (ODE). For functions of several variables there are partial differential equations (PDE). For functions of more than one variable we take partial derivatives instead of ordinary ones. The differential equations course (taken after Cal. III) is on ODEs. The eneral solution of a p th order ODE is any solution involvin p independent arbitrary constants. It is easy to verify that somethin is a solution to a differential equation; it is just a matter of takin derivatives and pluin into an equation. If a solution has the correct number of arbitrary constants then we can conclude that this is the eneral solution. The General of the Differential Equation With the analoy clearly stated, let us solve the second order ordinary differential equation 0 = d t + 2 2 x. Since the second derivatives of both sine and cosine are the neatives of themselves, it follows that cos t and sin t are solutions to our differential equation. Because the ODE has the simple properties (a homoeneous linear equation) that: it follows that (i) a constant times a solution is a solution and (ii) the sum of two solutions is a solution. x(t) = B cos t + C sin t is a solution, where B and C are arbitrary constants. Since we have solution to a second order ODE with two arbitrary constants, we can conclude that this is the eneral solution. We can interpret the constants B and C in terms of the initial position x 0 and the initial velocity v 0. x 0 = x(0) = B + C 0 = B This ives Another way of presentin this solution is d v(t) = x(t) = - B sin t + C cos t d t v 0 v 0 = v(0) = - B 0 + C C = v 0 x(t) = x 0 cos t + sin t. x(t) = A cos( t + ϕ) where the arbitrary constants are A, the amplitude, and ϕ, the phase anle. is called the anular frequency. This is related to the period T and the frequency f. The period is the time for one cycle. After one period the arument of the tri function shifts by. t t + T ( t + ϕ) ( t + ϕ) + This ives T =. The frequency is the number of cycles per time; since the time per cycle is the period we et f = /T. Combinin this ives T = f = and f = T =.
Chapter K - Oscillatory Motion 3 A x t 0 =- ϕ T= f = t -A Interactive Fiure K.2 - Enery Considerations The Enery of a Mass-Sprin System The total enery is the sum of kinetic and potential eneries. When the mass is at the turnin points x = ± A, its speed is zero; we can then write E = ( /2) k A 2. When it passes the equilibrium point x = 0 the potential enery is zero and thus its kinetic enery is the maximum; so must the speed be its maximum, v = ±v max. This allows us to write the enery with two equivalent forms for the total enery. E = 2 m v2 + 2 k x2 = 2 k A2 m v 2 2 max Equatin the two expressions for the total enery ives an expression for the maximum speed in terms of the amplitude. v max = k m A = A Speed and Position for General Simple Harmonic Motion For eneral simple harmonic motion we have Takin the time derivative of this ives the velocity. x(t) = A cos( t + ϕ) v(t) = - A sin( t + ϕ) Since both sine and cosine vary between ± we can identify the maximum speed as v max = A. This is equivalent to what we had for the mass-sprin case. The point here is to show that this is enerally true. If the mass-sprin enery is written in terms of the amplitude A then we can solve for v and, usin the mass-sprin value of, et v = ± A 2 - x 2. This expression is also enerally true for simple harmonic motion. To verify that enerally, we can write cos( t + ϕ) = x / A and sin( t + ϕ) = -v / ( A). Usin cos 2 + sin 2 = we can et the result. Example K. - Simple Harmonic Motion A particle moves in simple harmonic motion with a frequency of 3 Hz. At t = 0 the particle is released from rest from a distance of 2.4 cm from equilibrium.
4 Chapter K - Oscillatory Motion (a) What is the maximum speed and maximum acceleration of the particle? Since the particle is released from rest the amplitude is the initial distance from equilibrium. A = 2.4 cm = 0.025 m The anular frequency is related to the frequency f. f = 2 Hz f = 8.68 s - The maximum speed can now be found. v max = A =.96 m /s Since the acceleration is the second time-derivative of the position the acceleration follows from the differential equation. a = d t = 2-2 x The manitude of the maximum acceleration occurs at the larest value of x, which is the amplitude A. a max = 2 x = 60. m s 2 (b) At t = 0.22 s, what are the position, velocity and acceleration of the particle? Since at time zero the particle is at its maximum displacement we can conclude that the phase anle ϕ is zero. x(t) = A cos( t + ϕ) = A cos( t) Takin derivatives ives us the velocity and acceleration. At t 0 = 0.22 s we et d x d v v(t) = = - A sin( t) and a(t) = d t d t = -2 A cos( t) x(t 0 ) = 0.053 m, v(t 0 ) =.5 m /s and a(t 0 ) = -02 m s 2 (c) When the particle is.7 cm from equilibrium, then what are the speed and the manitude of the acceleration of the particle? The distance from equilibrium is x2. x2 =.7 cm = 0.07 m The speed is v2. v2 = A 2 - x 2 =.38 m /s From the discussion in part (a) we et the manitude of the acceleration. a2 = 2 x2 = 3 m s 2 K.3 - The Vertical Mass-Sprin
Chapter K - Oscillatory Motion 5 Interactive Fiure When a mass hans from a sprin we meed to add the effect of ravity. As before, the second law ives our differential equation. F net = m a The equilibrium position of this is when the forces cancel F net = 0. This ives We can redefine our coordinates relative to the new equilibrium position. If we insert this into our differential equation we et -k x + m = m d2 x k x eq = m. x = x + x eq -k x = m d2 x Here we have used the value of x eq and the fact that the derivatives of constants vanish, which implies the second derivative of x is the same as the second derivative of x. The interpretation of the above expression is simple. The effect of ravity is trivial. It just shifts the equilibrium position and we end up with simple harmonic motion about the new equilibrium position. d t 2. d t 2 Example K.2 - Vertical Mass/Sprin System When a mass is hun from a vertical sprin, the sprin stretches by 8.5 cm. What is the period of oscillation of this system? x eq = 8.5 cm = 0.085 m When hanin in equilibrium there are two forces actin on the mass, the sprin force k x eq actin upward and ravity downward. We do not know the mass m but we will see that it cancels; solve for k in terms of m. m k x eq = m k = From the anular frequency we can find the period. x eq k m = m x eq m = T = = x eq x eq 0.585 s
6 Chapter K - Oscillatory Motion The Physical Pendulum Axis θ d Center of Mass m θ Consider a riid body rotatin without friction about an axis. The center of mass is a distance d from the axis. At equilibrium, the center of mass will han below the center. Take the anle θ to be the anle of the line from the axis to the center of mass measured from vertical; note that θ = 0 is the equilibrium position. The only nonzero torque actin on the riid body is the torque due to ravity. This is τ net = τ rav = -m d sin θ. The direction of positive θ ives our sin convention for torque. The reason for the minus sin in the above expression is the torque tends toward smaller anles. Since the anular acceleration is the second time derivative of the anle, the rotational second law ives a differential equation. If we define τ net = I α -m d sin θ = I d2 d 2 m d d t θ = - 2 I sin θ d t 2 θ m d I then we et a differential equation of the form d 2 d t 2 θ = -2 sin θ. This is almost of the form of our simple harmonic motion equation d t 2 = - 2 x, except for the sine function. If we consider small anles then we et sin θ θ for small θ. We can then conclude that for small amplitude oscillations we have simple harmonic motion with an anular frequency iven by the expression above. Example K.3 - Physical Pendulum A uniform meter stick swins without friction about a perpendicular axis throuh the 20 cm line. What is its period of small oscillations? We use the parallel-axis theorem to find the moment of inertia. I = I cm + m d 2 = 2 m L2 + m d 2 m is not iven but will cancel; the other values are iven by L = m and d = 30 cm = 0.30 m. The period can then be found from the anular frequency.
Chapter K - Oscillatory Motion 7 m d I T = = I m d d = L2 2 + d2 =.53 s The Simple Pendulum The simple pendulum is a special case of the physical pendulum. It is the case where all the mass m is located at a point, the pendulum bob. If the bob is on the end of a strin of lenth L then we et Solvin for we et d = L and I = m L 2. L. Example K.4 - Texas A&M s Foucault Pendulum A Foucault Pendulum is a pendulum that precesses slowly in a circular path illustratin the earth s rotation underneath; this is due to the earth s Coriolis forces briefly mentioned in Chapter E. The Geore P. Mitchell Physics Buildin at Texas A&M University has a lare Foucault Pendulum with a period of 0.32 s. What is the lenth of this pendulum? L T = = L L = T 2 = 26.4 m