PHYS 1443 – Section 501 Lecture #25

1. PHYS 1443 – Section 501Lecture #25 Wednesday, April 28, 2004 Dr. Andrew Brandt • Simple Block-Spring System • Energy of the Simple Harmonic Oscillator • Pendulums • Damped oscillator PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

2. Announcements • The last HW on Ch. 12+13 is due tonight • No optional sections (*) in Ch 11-15 Explicitly, covered 11.1-5,7 12.1-6 13.1-8 14.1-5,7 • Last lecture will be Monday May 3 on Ch. 15 • On May 5, I will answer any questions that you may have on any of the material. If you have none it will be a short class—if you can use the time better with independent study that’s fine • Final exam will be Monday May 10 @5:30. It will be multiple choice only, so bring a Scantron form. It will be comprehensive, perhaps with a little more emphasis on ch 10-14. PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

3. Simple Block-Spring System A block attached at the end of a spring on a frictionless surface experiences acceleration when the spring is displaced from an equilibrium position. This becomes a second order differential equation If we denote The resulting differential equation becomes Since this satisfies condition for simple harmonic motion, we can take the solution see board Does this solution satisfy the differential equation? Let’s take derivatives with respect to time Now the second order derivative becomes Whenever the force acting on a particle is linearly proportional to the displacement from some equilibrium position and is in the opposite direction, the particle moves in simple harmonic motion. PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

4. More Simple Block-Spring System How do the period and frequency of this harmonic motion look? What can we learn from these? Since the angular frequency w is • Frequency and period do not depend on amplitude • Period is inversely proportional to spring constant and proportional to mass The period, T, becomes So the frequency is PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

5. Example for Spring Block System A car with a mass of 1300kg is constructed so that its frame is supported by four springs. Each spring has a force constant of 20,000N/m. If two people riding in the car have a combined mass of 160kg, find the frequency of vibration of the car after it is driven over a pothole in the road. Let’s assume that mass is evenly distributed on all four springs. The total mass of the system is 1460kg. Therefore each spring supports 365kg each. From the frequency relationship based on Hooke’s law Thus the frequency of vibration of each spring is How long does it take for the car to complete two full vibrations? The period is For two cycles PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

6. Example for Spring Block System A block with a mass of 200g is connected to a light spring for which the force constant is 5.00 N/m and is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.00 cm from equilibrium and released from rest. Find the period of its motion. From Hooke’s law, we obtain As we know, period does not depend on the amplitude or phase constant of the oscillation, therefore the period, T, is simply Determine the maximum speed of the block. From the general expression of the simple harmonic motion, the speed is PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

7. Since KE/PE E=KE+PE=kA2/2 -A A Energy of the Simple Harmonic Oscillator How do you think the mechanical energy of the harmonic oscillator look without friction? Kinetic energy of a harmonic oscillator is The elastic potential energy stored in the spring Therefore the total mechanical energy of the harmonic oscillator is Total mechanical energy of a simple harmonic oscillator is a constant for a particular motion and is proportional to the square of the amplitude Maximum KE is when PE=0 One can obtain speed PHYS 1443-501, Spring 2004 Dr. Andrew Brandt x

8. Example for Energy of Simple Harmonic Oscillator A 0.500kg cube connected to a light spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. a) Calculate the total energy of the system and the maximum speed of the cube if the amplitude of the motion is 3.00 cm. From the problem statement, A and k are The total energy of the cube is Maximum speed occurs when kinetic energy is the same as the total energy b) What is the velocity of the cube when the displacement is 2.00 cm. velocity at any given displacement is c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. Potential energy, PE Kinetic energy, KE PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

9. L q T m mg s The Pendulum A simple pendulum also performs periodic motion. The net force exerted on the bob is Since the arc length, s, is results If q is very small, sinq~q giving angular frequency Again a second degree differential equation, satisfying conditions for simple harmonic motion The period only depends on the length of the string and the gravitational acceleration The period for this motion is PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

10. Example for Pendulum Christian Huygens (1629-1695), a great clock maker, suggested that an international unit of length could be defined as the length of a simple pendulum having a period of exactly 1s. How long would our length unit be had this suggestion been followed? Since the period of a simple pendulum motion is The length of the pendulum in terms of T is Thus the length of the pendulum when T=1s is PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

11. O qmax P Torsion Pendulum When a rigid body is suspended by a wire to a fixed support at the top and the body is twisted through some small angle q, the twisted wire can exert a restoring torque on the body that is proportional to the angular displacement. The torque acting on the body due to the wire is k is the torsion constant of the wire Applying the Newton’s second law of rotational motion Then, again the equation becomes Thus, the angular frequency w is This result works as long as the elastic limit of the wire is not exceeded And the period for this motion is PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

12. Damped Oscillation More realistic oscillations involve an object losing its mechanical energy in time due to a retarding force such as friction or air resistance. Let’s consider a system whose retarding force is air resistance R=-bv (b is called damping coefficient) and restoration force is -kx The solution for the above 2nd order differential equation is Damping Term The angular frequency w for this motion is This equation of motion tells us that when the retarding force is much smaller than restoration force, the system oscillates but the amplitude decreases, and ultimately, the oscillation stops. We express the angular frequency as Where w0 is the natural frequency PHYS 1443-501, Spring 2004 Dr. Andrew Brandt

13. More on Damped Oscillation The motion is called Underdamped when the magnitude of the maximum retarding force Rmax = bvmax <kA How do you think the damping motion would change as the retarding force changes? As the retarding force becomes larger, the amplitude reduces more rapidly, eventually stopping at its equilibrium position Under what condition does this system not oscillate? The system is Critically damped Once released from non-equilibrium position, the object would return to its equilibrium position and stops. What do you think happens? If the retarding force is larger than the restoration force The system is Overdamped Once released from non-equilibrium position, the object would return to its equilibrium position and stop, but a lot slower than before PHYS 1443-501, Spring 2004 Dr. Andrew Brandt