Lecture 5 - Mech. Waves (Ch. 13.4-13.6)

# Lecture 5 - Mech. Waves (Ch. 13.4-13.6) - Lecture 5:...

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Unformatted text preview: Lecture 5: mechanical waves Sinusoidal wave function: Think about it for =0 at t=0 ---> y(x,t=0) = sin(kx ) y x Sinusoidal wave function: At time t=0 ---> y(x,t=0) = sin(x ) y x Now let k = 1/m and w = 2/s => y(x,t) = sin(x - 2t) t=0 x=2 x=- Sinusoidal wave function: At time t=0 ---> y(x,t=0) = sin(x ) y x At time t=1s ---> y(x,t=1) = sin(x - 0.5 ) Look at the dot in the figure At t=1, y = sin(- 0.5 )-- The same as for x=-0.5 when t was t=0 Let: k = 1/m and w = .5 /s => y(x,t) = sin(x - 0.5 t) t=0 x=2 x=- t=1 Sinusoidal wave function: Moves to the right at speed v Transverse Velocity and Acceleration o The general solution for a sinusoidal wave allows us to calculate the transverse (up and down) velocity and acceleration at any point and time: Example A wave function travels along a rope. A bead on the rope at x=10.0 moves up and down. What is the maximum speed of the bead? Example A wave function travels along a rope. A bead on the rope at x=10.0 moves up and down. What is the maximum speed of the bead? What sets the wave speed? Wave Speed :Tension & Density of cord o Tension on string exerts a net inward radial force: F r =2T sin 2T o Apply Newton: a r = F r /m = v 2 /R 2T /(2R )= v 2 /R o Speed of a transverse wave on a string: Higher tension, higher speed Heavier string, slower speed Example o A cord is 4 m long and has a mass of 0.1 kg. A transverse wave pulse makes a trip down the cord in 0.1 s. What is the tension in the cord? A. 40 N (left hand) B. 1 N (right hand) C. No idea (both hands) Example o A cord is 4 m long and has a mass of 0.1 kg. A transverse wave pulse makes a trip down the cord in 0.1 s. What is the tension in the cord? Energy and Waves o Waves can do work, so they must carry energy o Consider a sinusoidal wave on a string o Each mass element m acts like a harmonic oscillator Kinetic energy: K = ( m) v y 2 Using m= ( x): K = ( x) v y 2 Kinetic Energy o dK = dm v y 2 dK = (dx) v y 2 dK = [ A cos(kx- t)] 2 dx dK = 2 A 2 cos 2 (kx) dx (at t=0) o If we integrate the kinetic energy in a section one wavelength ( ) long, we find: K = 2 A 2 Total Energy...
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## This note was uploaded on 09/08/2008 for the course PHYS 3B taught by Professor Wu during the Spring '08 term at UC Irvine.

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Lecture 5 - Mech. Waves (Ch. 13.4-13.6) - Lecture 5:...

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