Phys 2101 Gabriela González

A wave “travels”, but the particles producing the wave don’t! Particles oscillate about a fixed mean position.

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1 Assume a sinusoidal wave travels in the x-direction. At each point in the x-axis, the transverse motion is in the y-direction, described by a function of time:

Wavelength: λ=2π/k Period: T=2π/ω Frequency: f=1/T=ω/2π 3 Velocity: v=ω/k=λ/T=λf

The equation of a transverse wave traveling along a very long string is given by y = 5.8 sin(0.015 x + 3.5 t), where x and y are expressed in centimeters and t is in seconds. Determine :

 the amplitude,  the wavelength  the frequency  the speed  the direction of propagation of the wave, and  the maximum transverse speed of a particle in the string.  What is the transverse displacement at x = 3.5 cm when t = 0.26 s?

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2 Write the equation for a wave traveling in the negative direction along the x axis and having an amplitude of 0.021 m, a frequency of 525 Hz, and a speed of 312 m/s.

ym = 0.021 m ω = 2πf = 3,300 rad/s k = ω /v =10.6 rad/m

y(x,t) = ym sin(ωt + kx) = 0.021 m sin(3,300 rad t /s +10.6 rad x /m)

A sinusoidal wave of frequency 475 Hz has a velocity of 250 m/s. € a) What is the phase difference between two displacements at a certain point at times 1.00 ms apart? Δφ = ωΔt = 2πfΔt = 3.0 rad

b) How far apart are two points that differ in phase by π/3 rad? € Δφ = kΔx = π /3 ⇒ Δx = π /3k k = 2πf /v ⇒ Δx = π /(6πf /v) = v /(6 f ) = 0.088 m = 8.8 cm 5

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For a string, we define: linear mass density µ = mass per unit length. If the string is udner tension τ, transverse waves can travel in the string with velocity v=(τ/µ)½ . • An object of mass M is suspended from the center of a string 10.0m long, with 8.00g/m linear density. The ends of the string are tied to walls 8 m apart. How much mass should be suspended from the string for pulses to travel at a velocity of 60.0 m/s?

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3 The elements of the string are moving up and down: they have kinetic energy. The tension of the string is acting as a restoring force to bring each element back to the center: each element has also potential energy.

The velocity of each element is

U=dy/ dt= d(ymsin(kx-ωt))/ dt=-ω ymcos(kx-ωt) The kinetic energy of each element dm=µ dx is 2 2 dK= ½ dm u = ½ µ dx (-ω ymcos(kx-ωt)) 2 2 2 2 = ½ µ dx ω y m cos (kx-ωt)) 2 2 2 2 dK/dt = ½ µ v ω y m cos (kx-ωt)) 2 2 Average rate of change in kinetic energy : µ v ω y m /4 Average rate of change in total energy (kinetic + potential)=

2 2 Average power transmitted by the wave: Pav= ½ µ v ω y m

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A string along which waves can travel is 2.50 m long and has a mass of 236 g. The tension in the string is 24.0 N. (a) What must be the frequency of traveling waves of amplitude 7.70 mm in order that the average power be 85.0 W? (b) If the average power carried by the string is 50W, and the frequency is kept constant, what will be the wave’s amplitude? (c) If the average power carried by the string is 50W, and the wave amplitude is kept constant, what will be the wave’s frequency?

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4 If we have two waves traveling in the same string, they will overlap and add up to a resultant wave. Overlapping waves do not in any way alter the travel of each other !!

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Two waves traveling in the same string, in the same direction, with the same frequency, will add up to a single resultant wave with the same frequency, but different amplitude and phase: y(x,t)= y1 sin(kx-ωt+φ1) + y2 sin(kx-ωt+φ2)

= ym sin(kx-ωt+φ) Interference term!

2 2 2 with ym =y1 +y2 +2y1y2cos(φ1-φ2)

and tan φ =(y1 sin φ1 + y2 sin φ2)/ (y1 cos φ1 + y2 cos φ2)

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5 2 2 2 Total amplitude: ym =y1 +y2 +2y1y2cos(φ1-φ2)

• If φ1-φ2=0, 2π,…, ym=y1+y2: constructive interference

• If φ1-φ2=π, 3π,…, ym=y1-y2: destructive interference

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Waves traveling in opposite directions can also be added. If they have the same frequency, they will produce a standing wave.

Waves in strings reaching the end of the string bounce back traveling in the opposite direction. If the wave can fit some integer number of wavelengths in the string, there is a standing wave, with λ=2L/n and f=n v/2L

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6 Strings A and B have identical lengths and linear densities, but string B is under greater tension than is string A. The figure shows four situations in which standing wave patterns exist on the two strings. In which situations is there the possibility that strings A and B are oscillating at the same resonant frequency?

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A nylon guitar string has a linear density of 7.0 g/m and is under a tension of 149 N. The fixed supports are 90 cm apart. The string is oscillating in the standing wave pattern shown in the figure. (a) What is this wave’s frequency? (b) What would be the frequency of the open string? (c) Which string is this one in the guitar? (d) Is the guitar tuned?

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7 Tempered scale • 82.4 E - open 6th string Major • 246.9 B - open 2nd string • 87.3 F scale: • 261.6 C - "middle C" • 92.5 F# • 277.2 C# • 264 Hz • 98.0 G • 293.6 D • 103.8 G# • 297 Hz • 311.1 D# • 110.0 A - open 5th string • 330 Hz • 329.6 E - open 1st string • 116.5 A# • 349.2 F • 352 Hz • 123.5 B • 370.0 F# • 130.8 C • 396 Hz • 392.0 G • 138.6 C# • 440 Hz • 415.3 G# • 146.8 D - open 4th string • 440.0 A - 5th fret on 1st string • 495 Hz • 155.6 D# • 466.1 A# • 164.8 E • 528 Hz • 493.8 B • 174.6 F • 523.2 C • 185.0 F# • 554.3 C# • 196.0 G - open 3rd string • 587.3 D • 207.6 G# • 622.2 D# • 220.0 A • 659.2 E - 12th fret on 1st string • 233.1 A#

15 http://entertainment.howstuffworks.com/guitar.htm

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