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Energy in Simple Harmonic Motion As a mass on a spring goes through its cycle of oscillation, energy is transformed from potential to kinetic and back to potential. Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 14-12 Energy: Kinetic & Potential Total Energy Either in terms of Kinetic K (x=0) Or Potential U (|x|= A) Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Total energy as 1) Kinetic energy in full E = m(vmax)2/2 or 2) Potential energy in full E= kA2/2 Equating these we get (vmax)2 = kA2/m But from UCM we found vmax = ωA Equating these two expressions yieds => Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Prob. 14-15 A block of mass m tied to a spring with spring constant k is undergoing simple harmonic oscillation. a)When the displacement of the mass is A/2, what fraction of the mechanical energy is kinetic energy and what fraction of it is potential energy? This is because total energy E = ½ kA2 b) At what displacement, as a fraction of A, is the energy half kinetic and half potential? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. P14-16: At maximum displacement position, x = A, a hammer hit gives the block a velocity v0 in the –x direction. >> Use Energy Conservation (often the easiest approach) a) Find Amplitude Α Note: At x = Α the kinetic energy is 0 kA2 = mv02 => A = (m /k) 1/2 v0 b) Find the velocity v2 at a point where x = A/2. 2 1 ⎞ 3⎛1 ⎞ 1 ⎛ A⎞ 1 1 1 1 ⎛1 2⎞ 1 1⎛1 mv22 + k ⎜ ⎟ = mv20 + 0 J ⇒ mv 22 = mv02 − ⎜ kA ⎟ = mv02 − ⎜ mv20 ⎟ = ⎜ mv20 ⎟ 2 ⎠ 2 ⎠ 4⎝2 ⎠ 2 ⎝2⎠ 2 2 2 4 ⎝2 4⎝2 ⇒ v2 = 3 v = 4 0 3 (0.40 m/s) = 0.346 m /s = 35 cm /s 4 Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Ch. 14: 7, 15,16, 22, 28, 36 ,44, 50 P44: Mass hanging at the end of spring in the gravitational field of Planet X is stretched by ΔL=31.2cm. • Pull mass down by x=10cm and released • Mass undergoes 10 oscillations in 14.5s • Find g 14-P50 Ultrasound Device m= 0.1g, f = 1MHz, max force F= 40,000 N, • Find max amplitude A • Find max speed v How to convert a snapshot graph to a history graph and vice versa? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Two more examples: Prob. 15.7 Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Prob. 15.9 Wave moving to the right at v=1m/s Given: History graph at x=0, Asked: snapshot at t=1s Think: same thing happening at x=0 at t= 1s has happened earlier at t= 0 sec read off value and place it at x= 1 or t= -1 sec at x= 2 And will happen at t=2 sec read off value and place it at x=-1 ! This is probably the toughest part in this chapter. We’ll work out more problems of this sort later Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Problem 15-8 Given snapshot graph at t=2s Draw history graph at x=0 for t=1 s to 8 s Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Prob.15-16 Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Ch15-P17 snapshot graph at t=0s Amplitude A= 4cm Wavelength λ=12m Frequency f = v / λ = 2.0 Hz Ch 15- P18 History Graph at x=0m of wave moving to the right at v=2m/s Amplitude A = 6cm Period T = 0.05 to 0.65 = 0.6 s Frequency f = 1/T = 1.33 Hz Wavelength λ = v/f = vT= 1.2 m Ch 15-28 Intensity of EM radiation •Sun emits EM waves with a power of 4x 1026W, what is the intensity of sunlight at the positions of Venus, Mars and Saturn? Intensity = Power / Area, A = 4πr2 Ch 15-P33 • •Source (loudspeaker) emits P= 35 W • •Receiver (microphone) located r= 50 m from source. (Microphone has area a=1cm2) – – – – Find sound intensity, intensity level at the microphone Intensity = Power / Area, where A= 4πr2 Intensity level: β = 10 dB log (Int / 10-12) = 90 db Added question: What is the total power received by the microphone? Power = Intensity x area, • Or, simpler way: P_mic / P_total = a / A Ch 15- 56 Given the form of the Wave function Can read off the Values of A, λ, T b) Calculate v c) After this put t=0.5 s, and x=0.2 m Into the expression for y(x,t) to get amplitude