Electric Potential - McMaster University

Electric Potential - McMaster University

Wave Motion II Sinusoidal (harmonic) waves Energy and power in sinusoidal waves For a wave traveling in the +x direction, the displacement y is given by y (x,t) = A sin (kx t) with = kv y A x -A Remember: the particles in the medium move vertically.

The transverse displacement of a particle at a fixed location x in the medium is a sinusoidal function of time i.e., simple harmonic motion: y = A sin (kx t) = A sin [ constant t] The angular angular frequency of the particle motion is ; the initial phase is kx (different for different x, that is, particles). = 2f =angular angular frequency radians/sec f =angular frequency cycles/sec

(Hz=hertz) Exampl e y A a e b -A c

d Shown is a picture of a travelling wave, y=A sin(kxt), at the instant for time t=0. Which particle moves according to y=A cos(t) ? ii) Which particle moves according to y=A sin(t) ? iii) If ye(t)=A cos(t+e ) for particle e, what is e ? Assume ye(0)=1/2A i) x Wave Velocity

The wave velocity is determined by the properties of the medium; for example, 1) Transverse waves on a string: v wave tension T mass/unit length (proof from Newtons second law and wave equation,

S16.5) 2) Electromagnetic wave (light, radio, etc.) 1 v c o o (proof from Maxwells Equations for E-M fields, S34.3) Example 1: What are , k and for a 99.7 MHz FM radio wave?

Example 2: A string is driven at one end at a frequency of 5Hz. The amplitude of the motion is 12cm, and the wave speed is 20 m/s. Determine the angular frequency for the wave and write an expression for the wave equation. Transverse Particle Velocities Transverse particle displacement, y (x,t) y v Transverse particle velocity, t

(x held y constant) this is called the transverse velocity (Note that vy is not the wave speed v ! ) v y a t t 2 Transverse acceleration, y

y 2 Standard Traveling sine wave (harmonic wave): y A sin( kx t ) y v y A cos(kx t ) t v y ay 2 A sin(kx t ) t 2 y

maximum transverse displacement, ymax = A maximum transverse velocity, vmax = A maximum transverse acceleration, amax = 2 A These are the usual results for S.H.M Example 3: y x string: 1 gram/m; 2.5 N tension vwave Oscillator: 50 Hz, amplitude 5 mm, y(0,0)=0 Find: y (x, t)

vy (x, t) and maximum transverse speed ay (x, t) and maximum transverse acceleration Solution Energy density in a wave ds dm dx Ignore difference between angular ds, angular dx (this is a good approx for small A, or large

): dm = dx ( = mass/unit length) Each particle of mass dm in the string is executing SHM so its total energy (kinetic + potential) is (since E= mv 2): 1 1 dE dm A dx A 2 2 2 2 2

2 The total energy per unit length is dE 1 A = energy density dx 2 2 2 Where does the potential energy in the string come from? Power transmitted by harmonic wave with wave speed v:

A distance v of the wave travels past a fixed point in the string in one second. So: P energy density v For waves on a string, power transmitted is P 12 2 A2 v Both the energy density and the power transmitted are proportional to the square of the amplitude and square of the frequency. This is a general property of sinusoidal waves. Example 4:

A stretched rope having mass per unit length of =5x10-2 kg/m is under a tension of 80 N. How much power must be supplied to the rope to generate harmonic waves at a frequency of 60 Hz and an amplitude of 6cm ?

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