Presentation on theme: "Moza M. Al-Rabban Professor of Physics"— Presentation transcript:
1 Moza M. Al-Rabban Professor of Physics firstname.lastname@example.org Physics 102 WavesLecture 2 More OscillationsMoza M. Al-RabbanProfessor of Physics
2 The Dynamics of SHMThe acceleration is proportional to the negative of the displacement at any time.
3 The Equation of Motion in SHM This is the equation of motion for the system. It is a homogeneous linear 2nd order differential equation.
4 Solving the Equation of Motion Q: How do you solve this differential equation?A: Our method - guess the solution, and see if it works.It works!
5 Example: Analyzing an Oscillator At t=0, a 500 g block oscillating on a spring is observed to be moving to the right at x=15 cm. At t=0.30 s, it reaches its maximum displacement of 25 cm.Draw a graph of the motion for one cycle.At what time in the first cycle is x=20 cm?The motion is SHM. The position equ. Of the block isThe object is initially moving to the right, which tells us that the phase constant must be between - and 0 rad. Thus
6 Example: Analyzing an Oscillator Draw a graph of the motion for one cycle.At what time in the first cycle is x=20 cm?The motion is SHM. The position equ. Of the block isThe block reaches its maximum displacement at time t = 0.30 s. At that instant of time
7 Vertical Oscillations Does the additional force of gravity change the motion?Now consider a mass m hanging from a spring with constant k. When the mass is attached, the spring stretches by DL.Finding DL is a static –equilibrium problem in which the upward spring force balances the downward weight force of the block. So,is a distance and it is a +ve number.is the displacement.In this configuration, the equilibrium position is given by:
8 Let the block oscillate around this equilibrium position. The origin of the y-axis at the block’s equilibrium positionIf the block moves upward, the spring gets shorter compared to its equilibrium length, but the spring is still stretched compared to its unstretched length.When the block is in position y, the spring is stretched by an amountAnd hence exerts an upward spring forceThe net force on the block at this point ismBecause, the motions are the same, everything we have learned about horizontal oscillation is equally valid for vertical oscillations.
9 Example 7: A Vertical Oscillator A 200 g block hangs from a spring with constant k = 10 N/m. The block is pulled down to a point where the spring is 30 cm longer than its unstretched length, then released.Where is the block and what is its velocity 3.0 s later?Although the spring begins by being stretched 30 cm, this is not the amplitude of the oscillation. Oscillations occur around the equilibrium position, so we have to begin by finding the equilibrium point where the block hangs motionless on the spring.Stretching the spring 30 cm pulls the block 10.4 cm below the equilibrium point,so A=10.4 cm.
10 Example 7: A Vertical Oscillator The initial conditionAt t= 3.0 s the block’s position is:The block is 7.4 cm above the equilibrium position, or 12.2 cm below the original end of the spring.Its velocity at this instant is
11 The PendulumThe pendulum’s position can be described by the arc of length s, which is zero when the pendulum hangs straight down.Because angles are measured ccw, s and are+ve when pendulum is to the right of center,-ve when it is to the left.
12 The equation of motion for an oscillating pendulum. The PendulumTwo forces are acting on the mass:The string tension and the weight.Using for acceleration “ around” the circle,Where the angle is related to the arc length by =s/L, soThe equation of motion for an oscillating pendulum.The sine function makes this equation more complicated than the oscillating spring’s ones.This is not a homogeneous linear 2nd order differential equation describing SHM.
13 The Small Angle Approximation The height of the triangle is h=r sin Suppose that the angle is “small”.The pendulum equation is:This is a homogeneous linear 2nd order differential equation describing SHM, but only for small swings (s<<L).
14 The Small Angle Approximation This is exactly the same equation as the equation for mass oscillating on a spring.The angular frequencyIs determined by the length of the string.
15 Example 8: A Pendulum Clock What length L must a pendulum have to give an oscillation period T of exactly 1 s?
16 Example 9: The Maximum Angle of a Pendulum A 300 g mass on a 30 cm long string oscillates as a pendulum. It has a speed of 0.25 m/s as it passes through the lowest point.What maximum angle does the pendulum reach?The speed of the lowest point is:
17 Conditions for SHMwe have solved all simple harmonic motion problems once we have solved the problem of horizontal spring.The restoring force of a spring, Fsp = -kx, is directly proportional to the displacement x from equilibrium.The pendulum’s restoring force, in the small angle approximation, is directly proportional to the displacement s.A restoring force that is directly proportional to the displacement from equilibrium is called a linear restoring force.For any linear restoring force, the equation of motion is identical to spring equation ( other than perhaps using different symbols).Any system with a linear restoring force will undergo simple harmonic motion around the equilibrium position.
18 Everything that we learn about an oscillating spring can be applied to the oscillations of any other linear restoring force, ranging from the vibration of airplane wings to the motion of electrons in electric circuits.Moreover, even when the restoring force is nonlinear (e.g., the pendulum), for sufficiently small displacements it can be treated as a linear system (small angle approximation).Therefore, we often treat atoms, molecules, and nuclei as small harmonic oscillators, even when the restoring forces involved may be very different from Hooke’s Law forces.
19 Identifying and Analyzing SHM If the net force acting on a particle is a linear restoring force, the motion will be simple harmonic motion (SHM) around the equilibrium position.The position x as a function of time is x(t) = A cos (wt + f0).The velocity v as a function of time is v(t) = - wA sin (wt + f0).The maximum speed is vmax = wA .The equations are given here in terms of x, but they can be written in terms of any other appropriate variables (y, q, etc.)The amplitude A and the phase constant f0 are determined by the initial conditions, through x0 = A cos f0 and v0x = -wA sin f0.The angular frequency w and the period T depend on the physics of the particular situation but are independent of A and f0.Mechanical energy is conserved. Thus, ½mvx2 + ½kx2 = ½kA2 = ½mvmax2. Energy conservation provides a relationship between position and velocity that is independent of time.
20 Clicker Question 1One person swings on a swing and finds that the period is 3.0 s. A second person of equal mass joins him on the same swing.With two people swinging, the period is:6.0 s;3.0 s;1.5 s;Between 3.0 s and 6.0 s;Can’t tell without knowing the length.
21 Damped Oscillations Drag force: b: the damping constant(damped oscillator)Damping factorSinusoidalvariationAmpl.Note that damping reduces the oscillation frequency.
22 Energy in Damped Systems The oscillator’s mechanical energy decays exponentially with time constant t.
23 Example: A Damped Pendulum A 500 g mass on a 50 cm long string oscillates as a pendulum. The amplitude of the pendulum is observed to decay to ½ of its initial value after 35 s.What is the time constant t of the damped oscillator?At what time t1/2 will the energy of the system have decayed to ½ of its initial value?
24 Clicker Question 2Which of these decays has the shortest time constant?
25 Driven Oscillations and Resonance Now, suppose we drive a damped mechanical oscillator with an external force F(t) = Fd cos(wdt). Then the equation of motion is:The system will show the property of resonance. The oscillation amplitude will depend on the driving frequency wd, and will have its maximum value when: i.e., when the system is driven at its resonant frequency w0.The width of the resonance curves depends on the amount of damping. Wider curves with smaller resonance curves correspond to more damping and larger values of b.
29 End of Lecture 2Before the next lecture, read Knight, Chapters 20.1 through 20.3.
30 Homework #1In taking your pulse, you count 75 heartbeats in 1 min. What are the period and frequency of your heart’s oscillations?What are the (a) amplitude,(b) frequency, and (c) phase constant of theoscillation shown in the Figure.The position of a 50 g oscillating mass is given by, where t is in s. Determine:a. The amplitude.b. The period.c. The spring constant.d. The phase constant.e. The initial conditions.f. The maximum speed.g. The total energy.h. The velocity at t = 0.4 s.
31 d. What is the first angle for which cos is less than 0.9900? A spring is hanging from the ceiling. Attaching a 500 g physics book to the spring causes it to stretch 20 cm in order to come to equilibrium.a. What is the spring constant?b. From equilibrium, the book is pulled down 10 cm and released. What is the period of oscillation?c. What is the book’s maximum speed? At what position or positions does it have this speed?21. Make a table with 4 columns and 8 rows. In row 1, label the columns (°), (rad), sin , and cos . In the left column, starting in row 2, write 0, 2, 4, 6, 8, 10, and 12.a. Convert each of these angles, in degrees, to radians. Put the results in column 2. Show 4 decimal places.b. Calculate the sines and cosines. Put the results, showing 4 decimal places, in columns 3 and 4.c. What is the first angle for which and sin differ by more than ?d. What is the first angle for which cos is less than ?e. Over what range of angles does the small-angle approximation appear to be valid?
32 29. A 205 g air-track glider is attached to a spring with spring constant 4.0 N/m. The damping constant due to air resistance is kg/s. The glider is pulled out 20cm from equilibrium and released. How many oscillations will it make during the time in which the amplitude decays toof its initial value?