How Does a Moving Speaker Affect Sound Frequency Perception?

In summary, a block with a speaker on the right side is placed on a table and connected to a spring with spring constant k. The block and speaker have a total mass of m and the amplitude of motion is A. The speaker emits sound waves with frequency f and speed of sound v. The highest and lowest frequencies heard by a person can be determined using the maximum and minimum speeds of the speaker, which are ±Aω. The sound level, measured in dB, is given by L=20log10(p)-β, where p and p0 are the pressures at two points and d is the distance between them. The distance d can be calculated using p0=λ/d and p=λ/(d+Δd
  • #1
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Homework Statement


A block with a speaker bolted to right side of it is put on the table. The left side of the block is connected to a spring having spring constant k and the block is free to oscillate in horizontal direction. The total mass of the block and speaker is m, and the amplitude of this unit's motion is A. The speaker emits sound waves of frequency f and speed of sound is v

aa-1.jpg


a. Determine the highest frequency heard by the person
b. Determine the lowest frequency heard by the person
c. If the maximum sound level heard by the person is β when the speaker is at its closest distance d from him, what is the minimum sound level heard by the person?

Homework Equations


maybe:
Doppler
T = 2π √(m/k)


The Attempt at a Solution


a. The highest frequency is when the speaker is the closest to the person. I am thinking using Doppler to find the frequency:
[tex]f_2=\frac{v±v_o}{v±v_s}f_1[/tex]

But the speed of the speaker is not constant so I don't think Doppler can be used.

The maximum speed is Aω = A√(k/m), but I am stuck...

b. Don't know
 
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  • #2
Well, it's asking for the maximum and minimum frequencies, so you only need the maximum and minimum speeds, which are ±Aω. That's parts a and b.
I'm assuming that sound level refers to the value measured in dB. That means that the sound level is given by [itex]L=20 \log_{10}\left(\frac{p}{p_0}\right)[/itex] which can be rearrange to be L=20log10(p)-β. Here p and p0 are the pressures at the two points. The pressure is inversely proportional to distance, so you can use p0=λ/d and p=λ/(d+Δd) get L as a function of β, d, and Δd. What is Δd?
 
  • #3
frogjg2003 said:
Well, it's asking for the maximum and minimum frequencies, so you only need the maximum and minimum speeds, which are ±Aω. That's parts a and b.
I'm assuming that sound level refers to the value measured in dB. That means that the sound level is given by [itex]L=20 \log_{10}\left(\frac{p}{p_0}\right)[/itex] which can be rearrange to be L=20log10(p)-β. Here p and p0 are the pressures at the two points. The pressure is inversely proportional to distance, so you can use p0=λ/d and p=λ/(d+Δd) get L as a function of β, d, and Δd. What is Δd?

I have never encountered the formula p0=λ/d. λ is the wavelength of the sound wave, and what is d?
 
  • #4
Oh, i used λ as an arbitrary constant. When you manipulate the math, it will be remove from the equations. d is the distance you were given in the diagram.
 
  • #5
frogjg2003 said:
Oh, i used λ as an arbitrary constant. When you manipulate the math, it will be remove from the equations. d is the distance you were given in the diagram.

Actually I still don't really understand the formula and the idea but let me try:
p0=λ/d and p=λ/(d+Δd), where Δd equals to 2A + d

L = 20 log (P/P0)
= 20 log P - 20 log p0
= 20 log λ - 20 log d - 20 log λ + 20 log (2d + 2A)
= 20 log 2 [(d + A)/d]

There is no β term

or

L = 20 log (P/P0)
= 20 log P - 20 log p0
= 20 log λ - 20 log d - β

The term λ does not cancel out

Please help
 
  • #6
I did another work and got answer like this:

[tex]minimum~sound~level = β + 20 log (\frac{d}{2A+d})[/tex]

Is this the correct answer?
 
  • #7
That looks right.
 
  • #8
OK thanks a lot for your help
 

1. What is a speaker attached to spring?

A speaker attached to spring is a type of audio device that uses a spring to produce sound. The speaker is connected to a spring, which is then stretched or compressed to create vibrations that produce sound waves.

2. How does a speaker attached to spring work?

The speaker attached to spring works by converting electrical energy into mechanical energy. The electrical signal from the audio source is sent to the speaker, which then moves the spring back and forth, creating sound waves that are amplified and projected through the speaker cone.

3. What are the benefits of using a speaker attached to spring?

One of the main benefits of using a speaker attached to spring is its ability to produce a wide range of frequencies, resulting in a clearer and richer sound. The use of a spring also allows for greater control over the quality and direction of the sound produced.

4. What are the common applications of a speaker attached to spring?

A speaker attached to spring is commonly used in audio devices such as headphones, speakers, and microphones. It is also used in musical instruments, such as guitars and pianos, to produce sound.

5. Are there any drawbacks to using a speaker attached to spring?

One potential drawback of using a speaker attached to spring is that it can be sensitive to external vibrations, which can affect the quality of the sound produced. Additionally, the use of a spring can make the speaker more prone to wear and tear over time.

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