3 Problems involving superposition

In summary: This will cause the wavelength to double (from 8.75 cm to 18 cm), and the wave will be in the second harmonic.
  • #1
mst3kjunkie
16
0
Here are the three problems I'm having trouble with:

Homework Statement


Two identical loudspeakers separated by distance d emit 170 Hz sound waves along the x-axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on. What are three possible values for d? Assume a sound speed of 340 m/s.


Homework Equations



phase difference = 2*Pi*(change in r/wavelength)+initial phase difference=2(m+1/2)Pi

interference is destructive if the path-length difference r=(m+1/2)wavelength

The Attempt at a Solution



find the wavelength

f=v/2L = v/wavelength
wavelength=v/f =340/170 =2

d=(m+1/2)wavelength

for m=0: d=1m
m=1: d=2m
m=2 d=3m


did I do this correctly?

Homework Statement


A steel wire is used to stretch a spring. An oscillating magnetic field drives the steel wire back and forth. A standing wave with three antinodes is created when the spring is stretched 8.0cm. What stretch of the spring produces a standing wave with two antinodes


Homework Equations



unsure

The Attempt at a Solution


the wave is in the third harmonic. The wavelength can be found using

wavelength=4L/m =4(.08)/3 = 8/75

I'm not sure how to get to the book's answer of 18 cm.

Homework Statement


Two loudspeakers 5.0m apart are playing the same frequency. If you stand 12.0 m in from of the plane of the speakers, centered between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 12.0m in front of them, you first hear a minimum of sound intensity when you are directly in from of one of the speakers.

a. what is the frequency of the sound? Assume a sound speed of 340 m/s
b. If you stay 12.0m in front of one of the speakers, for what other frequencies between 100 Hz and 1000 Hz is there a minimum sound intensity at this point?


Homework Equations


unsure for part b


The Attempt at a Solution


I've solved part a, f=170 Hz.

I don't even know how to start part b.
 
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  • #2
First problem:
mst3kjunkie said:
3. The Attempt at a Solution

find the wavelength

f=v/2L = v/wavelength
wavelength=v/f =340/170 =2

d=(m+1/2)wavelength

for m=0: d=1m
m=1: d=2m
m=2 d=3m


did I do this correctly?
Excessive busywork led you to the correct final equation, then you did the simple math wrong. plug in your m's again.
 
  • #3
2nd problem
mst3kjunkie said:
the wave is in the third harmonic. The wavelength can be found using

wavelength=4L/m =4(.08)/3 = 8/75

I'm not sure how to get to the book's answer of 18 cm.

You are looking for the stretch of the spring. Goto Hookes' Law. How does a change in tension affect the fundimental frequncy of a standing wave in a string?
 
  • #4
Isn't the [tex]\lambda[/tex] for the third harmonic [tex]\frac{2L}{3}[/tex]?

If you use hookes law, don't we need to know the spring constant k before using this information to solve this problem?
 
  • #5
You are right about the 2L/3, although this thread "died" 9 months ago.

But to answer your questions, you don't need to know the spring constant. You just need to know the proportionality between the stretch of the string, to the tension of the spring, to the speed of the wave in the string.

If the frequency is constant, by doubling the stretch of the spring, the tension will double, and the wave speed will increase by 1.41 (square root 2).

to have the same frequency go from being the 3rd harmonic to being the 2nd harmonic (2 is 2/3 of 3), the speed will have to increase by 3/2 of it's original, and the tension and therefore stretch will have to increase by the square root of 3/2Hmm, but that doesn't give 18 cm.

Just a minute...d'oh!

the stretch will increase by the square of 3/2 (not the root).
 
Last edited:

1. What is superposition?

Superposition is a principle in physics which states that when two or more waves interact, the total displacement at any given point is equal to the sum of the individual displacements caused by each wave.

2. How does superposition apply to problem-solving?

In problem-solving, superposition can be used to break down complex problems into smaller, simpler problems. By considering the contribution of each individual component or factor, we can then combine these solutions to find the overall solution to the problem.

3. What are some examples of problems involving superposition?

Some examples of problems involving superposition include analyzing the interference patterns of light waves, finding the electric potential at a point due to multiple charges, and determining the displacement of a particle in a vibrating string.

4. What is the difference between constructive and destructive interference in superposition?

In superposition, constructive interference occurs when the individual waves add up to create a larger amplitude at a given point, resulting in a stronger overall wave. Destructive interference, on the other hand, occurs when the individual waves cancel each other out, resulting in a weaker overall wave.

5. How can we use the principle of superposition to solve real-world problems?

Superposition can be applied to various real-world problems, such as analyzing the behavior of sound waves in a concert hall or designing antennas for better signal reception. By understanding how different waves interact and combine, we can optimize these systems for maximum efficiency.

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