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PHYS 1443 – Section 003 Lecture #21 Wednesday, Nov. 19, 2003 Dr. Mystery Lecturer 1. 2. 3. 4. 5. Fluid Dymanics : Flow rate and Continuity Equation Bernoulli’s Equation Simple Harmonic Motion Simple Block-Spring System Energy of the Simple Harmonic Oscillator Today’s Homework is #11 due on Wednesday, Nov. 26, 2003!! Next Wednesday’s class is cancelled but there will be homework!! Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 1 Flow Rate and the Equation of Continuity Study of fluid in motion: Fluid Dynamics If the fluid is water: Water Hydro-dynamics dynamics?? •Streamline or Laminar flow: Each particle of the fluid Two main types of flow follows a smooth path, a streamline •Turbulent flow: Erratic, small, whirlpool-like circles called eddy current or eddies which absorbs a lot of energy Flow rate: the mass of fluid that passes a given point per unit time m / t m1 1V1 1 A1l1 1 A1v1 t t t since the total flow must be conserved m1 m2 1 A1v1 2 A2v2 t t Equation of Continuity Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 2 Example for Equation of Continuity How large must a heating duct be if air moving at 3.0m/s along it can replenish the air every 15 minutes in a room of 300m3 volume? Assume the air’s density remains constant. Using equation of continuity 1 A1v1 2 A2v2 Since the air density is constant A1v1 A2v2 Now let’s call the room as the large section of the duct A2 v2 A2l2 / t V2 300 A1 0.11m 2 v1 v1 v1 t 3.0 900 Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 3 Bernoulli’s Equation Bernoulli’s Principle: Where the velocity of fluid is high, the pressure is low, and where the velocity is low, the pressure is high. Amount of work done by the force, F1, that exerts pressure, P1, at point 1 W1 F1l1 P1 A1l1 Amount of work done on the other section of the fluid is W2 P2 A2 l2 Work done by the gravitational force to move the fluid mass, m, from y1 to y2 is W3 mg y2 y1 Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 4 Bernoulli’s Equation cont’d The net work done on the fluid is W W1 W2 W3 P1 A1l1 P2 A2 l2 mgy2 mgy1 From the work-energy principle 1 1 2 mv2 mv12 P1 A1l1 P2 A2 l2 mgy2 mgy1 2 2 Since mass, m, is contained in the volume that flowed in the motion A1l1 A2 l2 Thus, and m A1l1 A2 l2 1 1 2 A2 l2 v2 A1l1v12 2 2 P1 A1l1 P2 A2 l2 A2 l2 gy2 A1l1 gy1 Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 5 Bernoulli’s Equation cont’d Since 1 1 A2 l2 v22 A1l1v12 P1 A1l1 P2 A2 l2 A2 l2 gy2 A1l1 gy1 2 2 We obtain 1 2 1 2 v2 v1 P1 P2 gy2 gy1 2 2 Reorganize P1 1 2 1 2 Bernoulli’s v1 gy1 P2 v2 gy2 Equation 2 2 Thus, for any two points in the flow 1 2 P1 v1 gy1 const. 2 For static fluid P2 P1 g y1 y2 P1 gh 1 2 2 P P v v 2 1 1 2 For the same heights 2 Pascal’s Law The pressure at the faster section of the fluid is smaller than slower section. Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 6 Example for Bernoulli’s Equation Water circulates throughout a house in a hot-water heating system. If the water is pumped at a speed of 0.5m/s through a 4.0cm diameter pipe in the basement under a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter pipe on the second 5.0m above? Assume the pipes do not divide into branches. Using the equation of continuity, flow speed on the second floor is 2 A1v1 r12 v1 0.020 v2 0.5 1.2m / s 2 A2 r2 0.013 Using Bernoulli’s equation, the pressure in the pipe on the second floor is 1 P2 P1 v12 v22 g y1 y2 2 1 5 3.0 10 1 103 0.52 1.22 1 103 9.8 5 2 2.5 105 N / m 2 Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 7 Simple Harmonic Motion What do you think a harmonic motion is? Motion that occurs by the force that depends on displacement, and the force is always directed toward the system’s equilibrium position. What is a system that has such characteristics? A system consists of a mass and a spring When a spring is stretched from its equilibrium position by a length x, the force acting on the mass is F kx It’s negative, because the force resists against the change of length, directed toward the equilibrium position. From Newton’s second law F ma kx we obtain a k x m Condition for simple k This is a second order differential equation that can d 2 x x harmonic motion be solved but it is beyond the scope of this class. m dt 2 Acceleration is proportional to displacement from the equilibrium What do you observe Acceleration is opposite direction to displacement from this equation? Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 8 Equation of Simple Harmonic Motion d 2x k x 2 dt m The solution for the 2nd order differential equation x A cos t Amplitude Phase Angular Frequency Phase constant Generalized expression of a simple harmonic motion Let’s think about the meaning of this equation of motion What happens when t=0 and =0? What is if x is not A at t=0? x A cos0 0 A x A cos x' cos 1 x' What are the maximum/minimum possible values of x? Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu A/-A An oscillation is fully characterized by its: •Amplitude •Period or frequency •Phase constant 9 More on Equation of Simple Harmonic Motion What is the time for full cycle of oscillation? The period Since after a full cycle the position must be the same x Acos t T A cost 2 T How many full cycles of oscillation does this undergo per unit time? 2 One of the properties of an oscillatory motion f 1 T 2 Frequency Let’s now think about the object’s speed and acceleration. What is the unit? 1/s=Hz x Acost dx Asin t Max speed v max A dt Max acceleration dv 2 2 Acceleration at any given time a A cos t x a 2 A dt max What do we learn Acceleration is reverse direction to displacement about acceleration? Acceleration and speed are /2 off phase: When v is maximum, a is at its minimum Speed at any given time Wednesday, Nov. 19, 2003 v PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 10 Simple Harmonic Motion continued Phase constant determines the starting position of a simple harmonic motion. x Acost x t 0 A cos At t=0 This constant is important when there are more than one harmonic oscillation involved in the motion and to determine the overall effect of the composite motion Let’s determine phase constant and amplitude xi A cos At t=0 vi A sin vi By taking the ratio, one can obtain the phase constant tan xi 1 By squaring the two equation and adding them together, one can obtain the amplitude A cos sin 2 2 Wednesday, Nov. 19, 2003 2 2 x 2 vi A i xi2 A2 cos 2 vi2 2 A2 sin 2 2 A PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu vi 2 xi 2 11 Example for Simple Harmonic Motion An object oscillates with simple harmonic motion along the x-axis. Its displacement from the origin varies with time according to the equation; x 4.00mcost where t is in seconds and the angles is in the parentheses are in radians. a) Determine the amplitude, frequency, and period of the motion. 4 . 00 m cos t Acos t x A 4.00m The angular frequency, , is From the equation of motion: The amplitude, A, is Therefore, frequency and period are T 2 2 2s f 1 1 s 1 T 2 b)Calculate the velocity and acceleration of the object at any time t. Taking the first derivative on the equation of motion, the velocity is By the same token, taking the second derivative of equation of motion, the acceleration, a, is Wednesday, Nov. 19, 2003 dx 4 . 00 sin t v m / s dt d 2 x 4.00 2 cos t m / s 2 a 2 dt PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 12 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. a k x m This becomes a second order differential equation If we k d 2x k x denote 2 m dt m 2 d x The resulting differential equation becomes x 2 dt Since this satisfies condition for simple x A cost harmonic motion, we can take the solution Does this solution satisfy the differential equation? Let’s take derivatives with respect to time dx A d cost sin t Now the second order derivative becomes dt dt d d 2x sin t 2 cost 2 x 2 dt dt 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. Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 13 More Simple Block-Spring System How do the period and frequency of this harmonic motion look? Since the angular frequency is The period, T, becomes So the frequency is Special case #1 x cost f T 2 2 1 1 T 2 2 k m m k k m What can we learn from these? •Frequency and period do not depend on amplitude •Period is inversely proportional to spring constant and proportional to mass Let’s consider that the spring is stretched to distance A and the block is let go from rest, giving 0 initial speed; xi=A, vi=0, 2 dx d x v sin t a 2 2 cos t ai 2 kA/ m dt dt This equation of motion satisfies all the conditions. So it is the solution for this motion. Suppose block is given non-zero initial velocity vi to positive x at the instant it is at the equilibrium, xi=0 Is this a good vi 1 1 tan tan x cos t A sin t solution? x Special case #2 i Wednesday, Nov. 19, 2003 PHYS 1443-003, Fall 2003 Dr. Jaehoon Yu 14