Standing waves - University of Toledoastro1.panet.utoledo.edu/~mheben/PHYS_2130/Chapter17-1_mh.pdf · Standing waves and resonance For a string clamped at both ends, at certain frequencies,

Standing waves The interference of two sinusoidal waves of the same frequency and amplitude, travel in opposite direction, produce a standing wave.

y1(x, t) = ymsin(kx - ωt), y2(x, t) = ymsin(kx + ωt)

resultant wave: y'(x, t) = [2ymsin kx] cos(ωt)

nodes Anti-nodes

y

kx

0 π/2 π 3π/2 2π

y'(x, t) = [2ymsin kx] cos(wt)

If kx = nπ (n = 0, 1, …), we have y' = 0; these positions are called nodes. x = nπ/k = nπ/(2π/λ) = n(λ/2)

y

kx

0 π/2 π 3π/2 2π

If kx = (n + ½)π (n = 0, 1, …), y'm = 2ym (maximum); these positions are called antinodes, x = (n + ½ )(λ/2)

y'(x, t) = [2ymsin kx] cos(ωt)

Standing waves and resonance For a string clamped at both ends, at certain frequencies, the interference between the forward wave and the reflected wave produces a standing wave pattern. The string is said to resonate at these certain frequencies, called resonance frequencies. L = l/2, f = v/λ = v/2L 1st harmonic or fundamental mode

L = 2(λ/2 ), f = 2(v/2L) 2nd harmonic

f = n(v/2L), n = 1, 2, 3… nth harmonic

Reflection at a Boundary

Case a) String tied to wall. – The reflected and incident pulses must have opposite signs. A node is generated at the end of the string.

Case b) String end can move. – The reflected and incident pulses must have same sign. An antinode is generated at the end of the string.

a b

a b Reflection at a Boundary



a b Reflection at a Boundary



Sample Problem

A pulse is traveling down a wire and encounters a heavier wire. What happens to the reflected pulse?

1.  The reflected wave changes phase 2.  The reflected wave doesn’t change phase 3.  There is no reflected wave 4.  none of the above

Problem 16-84

Note y1 and y2 are cosines, not sines – but y’ is *still* = y1 + y2

where α = πx ‒ 4πt and β = πx + 4πt.

A standing wave results from the sum of two transverse traveling waves given by: y1(x, t) = 0.050 cos(πx - 4πt)

y2(x, t) = 0.050 cos(πx + 4πt) where x, y1, and y2 are in meters and t is in seconds.

Alternatively, cos(α) = sin(α + π/2) = sin(α)cos(π/2) + sin(π/2)cos(α)

(trig identity)

Problem 16-84 A standing wave results from the sum of two transverse traveling waves given by: y1(x, t) = 0.050 cos(πx ‒ 4πt)


where α = πx ‒ 4πt and β = πx + 4πt.



Amplitude in space Amplitude in time



A) Find smallest non-negative x where y’=0.

Amplitude of composite wave: 2ymcos(πx) = 0 => πx = π/2, 3π/2, 5π/2, ...

Smallest when x = 1/2

Problem 16-84 A standing wave results from the sum of two transverse traveling waves given by: y1(x, t) = 0.050 cos(πx - 4πt)


B) Find times the particle at x=0 has zero velocity. Take the temporal derivative of y’:

sin(4πt) = 0 => 4πt = 0, π, 2π, 3π, ... => t = 0, 1/4, 1/2, 3/4, ... x=0 =1

Pipes

{ Open on both ends

{Open on one end, closed on the other end

Antinodes both ends.

node on closed end, antinode on open end.

Chapter 17 Waves (Part II)

Sound wave is a longitudinal wave.

The speed of sound Speed of a transverse wave on a stretched string

The speed of sound Speed of a sound wave (longitudinal):

Speed of sound in gases & liquids is described by the bulk modulus B ( = - Δp/(ΔV/V)) and density ρ:

Speed of sound:

air(0oC): 331 m/s, air (20oC): 343 m/s, water (20oC): 1482 m/s

The speed of sound

Speed of sound in a solid is described by Young’s modulus E and density ρ:

Speed of sound in aluminum: E = 7x1010 Pa, ρ = 2.7x103 kg/m3 => v = 5100 m/s,

In general, vsolid > vliquid > vgas

Speed of a sound wave (longitudinal):

Traveling Sound Wave To describe the sound wave, we use the displacement of an element at position x and time t:

s(x, t) = smcos( kx – ωt )

sm: displacement amplitude

k = 2π/λ ω = 2π/T = 2πf

As the wave moves, the air pressure at each point changes:

Δp(x, t) = Δpmsin( kx – ωt )

Pressure amplitude: Δpm = (vρω)sm

Interference Two sound waves traveling in the same direction, how they would interfere at a point P depends on the phase difference between them.

If the two waves were in phase when they were emitted, then the phase difference depends on path length difference: ΔL = L1 – L2:

ϕ/2π = ΔL/λ

ΔL = L1 – L2

Interference

Δϕ/2π = ΔL/λ

If Δϕ = m (2π) or ΔL/λ = m ( m = 0, 1, 2 …)

We have fully constructive interference ΔL = d sinθ = mλ

If Δϕ = (2m +1)π or ΔL/λ = m + ½ ( m = 0, 1, 2 …)

We have fully destructive interference d sinθ = (m + ½) λ

For distant points,

ΔL

ΔL = d sinθ d θ

ΔL = d sinθ = mλ

D

y

y/D = tanθ

Problem 17-19 Two loudspeakers are located 3.35m apart. A listener is 18.3m from one and 19.5m from the other speaker.

A) What is the lowest audible frequency (20 Hz – 20 kHz) that gives a minimum (fully destructive interference) signal at the listener’s location?

ΔL

d θ D

y

L1=18.3m, L2=19.5m => ΔL = 1.2m

We want the path difference to be an odd multiple of the half wavelength. Or in other words,

ΔL = (m+1/2)λ, where m=0, 1, 2, ...


A) What is the lowest audible frequency (20Hz - 20kHz) that gives a minimum signal at the listener’s location?

ΔL

d θ D

y

L1=18.3m, L2=19.5m => ΔL = 1.2m

We want ΔL = (m+1/2)λ, where m=0, 1, 2, ...

vsound = f λ => flowest happens when λ is the largest


A) What is the lowest audible frequency (20Hz - 20kHz) that gives a minimum signal at the listener’s location?

DL

d q D

y

L1=18.3m, L2=19.5m => DL = 1.2m

We want DL = (n+1/2)l, where n=0, 1, 2, ...


B) What is the second lowest audible frequency (20Hz - 20kHz) that gives a minimum signal at the listener’s location?

DL

d q D

y

L1=18.3m, L2=19.5m => ΔL = 1.2m

We want ΔL = (n+1/2)λ, where n=0, 1, 2, ...


D) What is the lowest audible frequency (20Hz - 20kHz) that gives a maximum signal at the listener’s location?

DL

d q D

y

L1=18.3m, L2=19.5m => ΔL = 1.2m

We want ΔL = nλ, where n=1, 2, ...

Sample Problem

1) 4.1km 2) 1200m 3) 343m 4) Yikes, head for cover 5) none of the above

Today, a lightning bolt is seen in the distance. You start counting when you saw the flash. Twelve seconds elapsed until you heard the thunder. How far away was the lightning?

Daily Quiz-Lightning, March 27, 2006

Today, a lightning bolt is seen in the distance. You start counting when you saw the flash. Twelve seconds elapsed until you heard the thunder. How far away was the lightning?

1) 4.1km 2) 1200m 3) 343m 4) Yikes, head for cover 5) none of the above

d = vsoundt = (343m/s)(12s) = 4.1km = 2.5miles

Intensity and Sound Level

Spherical surface area: 4πr2

Intensity of a sound wave at a surface is defined as the average power per unit area

I = P/A For a wave s(x, t) = smcos( kx – ωt )

I = ½ ρvω2s2m v = ω/k

Variation of intensity with distance.

A point source with power Ps

I = Ps/4πr2

(Watts/m2)

Ps – a property of the source (W, or J/s)

The figure indicates three small patches 1, 2, and 3 that lie on the surfaces of two imaginary spheres; the spheres are centered on an isotropic point source S of sound.

A) Rank the patches according to the intensity of the sound on them, greatest first.

B) If the rates at which energy is transmitted through the three patches by the sound waves are equal, rank the patches according to their area, greatest first.

I = Ps/4πr2 r1 = r2 => I1 = I2

r3 > r1 => I3 < I1

I = Ps/A r3 > r1 => A3 > A1

r1 = r2 => A1 = A2

The decibel Scale The deci(bel) scale is named after Alexander Graham Bell

Sound level b is defined as

β = (10 dB)log (I/I0)

Unit for β: dB(decibel) I0 is a standard reference intensity, I0 = 10-12 W/m2

Threshold of hearing: 10-12W/m2

Threshold of pain (depends on the type of music): 1.0W/m2

An increase of 10 dB => sound intensity multiplied by 10.

• The decibel scale – Sound intensity range for human ear: 10-12 – 1 W/m2

– More convenient to use logarithm for such a big range.

– Sound level b is defined as

b = (10 dB)log10(I/I0)

• Unit for b: dB(decibel)

I0 is a standard reference intensity, I0 = 10-12 W/m2

Examples: at hearing threshold: I = I0, b = 10 log101 = 0, normal conversation: I ~ 106 I0, b = 10 log10106 = 60 dB Rock concert: I = 0.1 W/m2, b = 10 log10(0.1/10-12) = 110dB

Beats Two sound waves with frequencies f1 and f2 ( f1 ~ f2 ) reach a detector, the intensity of the combined sound wave vary at beat frequency fb

fb = f1 – f2

The Doppler effect When the source or the detector is moving relative to the media, a frequency which is different from the emitted frequency is detected.

This is called Doppler effect.

• Case 1: source stationary, the detector is moving towards the source:

Note: f ' is greater than f.

The detected frequency is:

Case 2: source stationary, detector moving away from the source:

f ' is smaller than f.

Cases 3 & 4: source moving, detector stationary

Source is moving toward detector, “-” sign, f ' > f

Source is moving away from detector, “+” sign, f ' < f

General formula for Doppler effect v: speed of sound vD: detector speed vS: source speed

Rules:

Moving towards each other: frequency increases

Moving away from each other: frequency decreases

For numerator,

if detector is moving towards the source, “ + ”

if detector is moving away from the source, “ - ”

For denominator,

if source is moving away from the detector, “ + ”

if source is moving towards the detector, “ - ”

Standing waves - University of Toledoastro1.panet.utoledo.edu/~mheben/PHYS_2130/Chapter17-1_mh.pdf · Standing waves and resonance For a string clamped at both ends, at certain frequencies,

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