THE VIBRATIONS OF MOLECULES # I THE SIMPLE HARMONIC OSCILLATOR

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1 THE VIBRATIONS OF MOLECULES # I THE SIMPLE HARMONIC OSCILLATOR STUDENT INSTRUCTIONS by George Hardgrove Chemistry Department St. Olaf College Northfield, MN hardgrov@lars.acc.stolaf.edu These instructions accompany the MATHCAD file horun.mcd. Copyright George Hardgrove All rights reserved. You are welcome to use this document in your own classes but commercial use is not allowed without the permission of the author. Goal: To explore the mathematical representation of classic harmonic motion and the effect of mass and force constant on this motion. Objectives: Students will be able to: 1. Derive the equations of motion for a harmonic oscillator from the Lagrangian for this oscillator;. Explain the effect of k, the force constant, and m, the mass, on the frequency of the vibration of an oscillator. Prerequisites: 1. Mathcad 6.0+ is required to use this document;. Elementary physics concepts: the first and second derivative of position with respect to time representations of velocity and acceleration; 3. Differential calculus: take the derivative of simple functions; and do simple partial derivatives; 4. Only entry level skill with Mathcad is required to use the program. More advanced skills are required to create a document like this one. page 1

2 This program illustrates the classical harmonic oscillator. First we will derive the differential equation of motion for a mass m connected to a heavy wall with a spring of force constant k. The coordinate system is chosen so that when the mass is at rest with no forces acting on it the value of x = 0. When the spring is stretched x > 0, and when the spring is compressed x< 0. Draw a sketch of the physical system described here. The potential energy for this system is: V = 1 kx (1) The kinetic energy T is given by the expression T = 1 mx& () where &x is the time derivative of the distance x, i.e. the velocity. In the potential energy V, k is the force constant and x is the displacement from the equilibrium position. As you can see from equation (1) a larger x gives a greater potential energy. Since the force on the object is a negative gradient of the potential energy then the larger the value of x the greater the restoring force that pulls the object back to the equilibrium position. To obtain the differential equation of motion we use the Lagrangian L 1 1 L = T V = mx& k x (3) which is then substituted into Lagrange's equation of motion and gives the differential equation d d t L L x& x = 0 (4) k &&x + m x = 0 or d x k + d t m x = 0 (5) page

3 EXERCISE 1 Substitute L in equation (3) into Lagrange's equation (4), perform the required steps to obtain differential equation (5). Remember that when you take a partial derivative you assume that all other variables are constants. EXERCISE In the computational document set k=3, m=1, and in the initial values matrix set x=. and dx/dt = 0. Run the program (press F9) and note the graph on page. The red line represents x as a function of time and the green dash line represents dx/dt. When x=0 what is the velocity dx/dt? When the dx/dt=0 what is the value of x? EXERCISE 3 Now change k by a factor of 10. How has the frequency changed? Return k to its original value and raise m by a factor of 10. What happens to the frequency? EXERCISE 4 The amplitude is defined as the maximum displacement from equilibrium, for example in Exercise the amplitude is 0.. How do changes in mass and force constant affect the amplitude of the oscillation? EXERCISE 5 Change k to 300 and the mass m to 1. Now observe the graph. Is this the expected behavior? Is there an explanation for this graph? page 3

4 THE DIATOMIC MOLECULE The same type of differential equation applies to a diatomic molecule with modest adjustments of the above equations. We describe the motions in terms of an internal coordinate q which is the amount that the distance between the atoms has been distorted from the equilibrium bond length. For the mass we must use the reduced mass which is calculated from the individual atom masses as follows: m1 m µ = m + m 1 ( 6a) where m 1 and m are the masses in kg of the individual atoms. This quantity can be calculated using gram molecular weights µ = M1M 1 1 M1 + M N av 10 3 (6b) N av is Avogadro's number. The kinetic energy is T = 1 µ q& ( 7) The potential energy is written as V = 1 k q (8) The differential equation of motion then becomes &&q k + q = 0 (9) µ page 4

5 where d q &&q = d t (10) A figure suitable for animation is given at the end of the computational document. Place the cursor just to the right of this diagram and then click on Window then Animation, the create/playback dialog box should appear. Click on create. Set the parameters for the animate mode to From = 0, To = 30, At 10 frames per second. Next move the cursor to the upper-left side of the diagram and hold down the left mouse button to mark off the region you wish to animate. Include in this region the entire diagram. Be sure that Mathcad is in the automatic recalculate mode. It is in this mode if the light bulb in the tool bar is depressed.. Now click on animate and the computer will prepare the frames of the atoms at different positions. When the calculations are complete a small playback picture appears at the left. Click on the arrow at the left bottom and the animation will play back. The number of frames that you will be able to create and the speed at which the animation is created depends on the speed of the cpu in the computer at which you are working. page 5

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