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PHY 2048

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  1. PHY 2048 • Lecture 8

  2. CHAPTER 14 Periodic Motion

  3. velocity vector is perpendicular to the position vector acceleration vector points inward Position vector is speed is Velocity vector is acceleration vector is magnitude of acceleration is uniform circular motion - summary (slide 4, lecture 3) Periodic motion x(t) = R cos (ωt) R: Amplitude (maximum displacement) ω: Angular frequency (radians per second) ω = 2πf f: frequency (number of cycles in unit time) T=1/f (period: time for one cycle) centripetal acceleration

  4. Simple Harmonic Motion T/2 A (3T)/2 (5T)/2 (7T)/2 (9T)/2 T/4 period (3T)/4 Angular frequency frequency x(t) t 3T 4T 5T T 2T -A x(t) = A cos (ωt) = A cos (2πft) = A cos (2πt/T) vx(t) = -Aω sin(ωt) ax(t) = -Aω2 cos(ωt) ax(t) = -ω2x(t) Fx = -mω2x Fx=-kx with k>0 produces simple harmonic motion ω2=k/m

  5. Analyzing a harmonic motion plot It is evident from the graph that the object is moving in the positive x direction from A to C and in the negative x direction from C to E. The object is in its equilibrium position at A, C and E. The slope of the graph is positive from A to B and from D to E. The velocity is positive in these two intervals. The slope of the graph is negative from B to D and the velocity is negative. One can also say that the velocity is zero at B and D. In order to make statements about acceleration in simple harmonic motion, it is best to use the fact that the a(t) is opposite of x(t) with an additional multiplicative factor. The acceleration is positive from C to E and negative from A to C. The acceleration is zero at A, C and E. x(t) = A sin(ωt) vx(t) = Aω cos(ωt) ax(t) = -Aω2 sin(ωt)

  6. An example of a simple harmonic motion Block is released at rest from x=A The spring is stretched at this point The spring pulls the block toward its equilibrium position (x=0) The block gains kinetic energy till it reaches x=0 Because the block has kinetic energy, it compresses the spring till it reaches x=-A Because the spring is compressed at x=-A, it pushes the block out to its equilibrium position The block gains kinetic energy till it reaches x=0 Because the block has kinetic energy, it stretches the spring till it reaches x=A This is one full cycle and we have simple harmonic motion Let the period (time for one cycle) be T= 0.1 s The frequency f = 1/T = 10 s-1 = 10 Hertz (short form is Hz) The block goes through 10 cycles in one second

  7. m/k = (L0-L)/(2g) period A crate is kept on a spring with spring constant k and uncompressed length L0 The crate is let go and the spring compresses to a final length of L before the crate bounces back How long does it take for the crate to come back again to the equilibrium position? Let m be the mass of the crate. The crate has no kinetic energy when it starts at equilibrium position or when it goes down to the spring’s fully compressed position The gravitational potential energy lost in going from L0 to L is mg(L0-L) The elastic potential energy gained in going from L0 to L is 1/2 k(L0-L)2 Since total mechanical energy is conserved and there is no kinetic energy at L or L0 mg(L0-L) = 1/2 k (L0-L)2 The time taken for the crate to go from L0 to L and back to L0 is one period The period can be found if we know L0 and L.

  8. Given the initial position and velocity, use Initial velocity to find the phase angle ϕ to find the amplitude, A Finally, use and Given m and k, use to find the angular frequency, ω Phase in simple harmonic motion ϕ is called the phase angle Initial position

  9. How to find the phase from a plot of x(t) We know that the amplitude A=M. We know that the period is T. The angular frequency is ω=2π/T The standard form for x(t) is x(t) = M cos(ωt+ϕ) We know that x(-N) = M from the plot ω(-N)+ϕ=0 ϕ=ωN=(2πN)/T

  10. Energy in simple harmonic motion The force, Fx=-kx, responsible for simple harmonic motion is like a spring force This is a conservative force and we can define U(x) = 1/2 kx2 as the potential energy Kinetic energy Potential energy Total mechanical energy is a constant

  11. θ=0 In radians x Frequency Angular frequency Period The simple pendulum Assume θ is very small Motion of the pendulum is almost horizontal θ=θmax We have simple harmonic motion with k=mg/L Angular frequency of the pendulum does not depend on mass

  12. Torque about the pivot point O is Angular frequency frequency Newton’s law for rotational motion period The physical pendulum Assume θ is very small We have simple harmonic motion in the variable θ

  13. N(t) mg A cool way to measure g To measure the magnitude of the acceleration due to gravity g in an unorthodox manner, a student places a ball bearing on the concave side of a flexible speaker cone . The speaker cone acts as a simple harmonic oscillator whose amplitude is A and whose frequency f can be varied. The student can measure both A and f with a strobe light. Take the equation of motion of the oscillator as y(t) = A cos(ωt+ϕ), where ω=2πf and the y axis points upward. Let us assume that the ball bearing is always in contact with the speaker surface. There must be a normal force acting on the ball due to the speaker surface Since the ball bearing is always in contact with the speaker surface, its acceleration must be the same as that of the speaker surface. The minimum N(t) reaches is m(g-4π2f2A) N(t) must be always larger than zero for the ball to be in contact with the speaker surface. Let fb be the critical frequency at which frequency and above, the ball loses contact with the speaker surface. Then g=4π2fbA.