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### Transcript of Standing waves - University of mheben/PHYS_2130/Chapter17-1_mh.pdf · PDF file Standing...

• 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

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 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.

• 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)

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

where α = πx ‒ 4πt and β = πx + 4πt.

• 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)

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

Amplitude in space Amplitude in time

• 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)

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

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)

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

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, ...

• 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 (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

• 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 (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, ...

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

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, ...

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

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