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### Sounds Of Music C

Posted: September 5th, 2018, 5:54 pm

A speaker produces a sound with an intensity of 1 x 10^-4 W/m^2. What is the intensity, in decibels, of the sound produced by 4 of these speakers?

### Re: Sounds Of Music C

Posted: September 5th, 2018, 6:01 pm
`10^-4 W/m^2 is 80dB. A factor of 4 is an additional 6dB, so the configuration has a total sound intensity of 86dB (assuming only constructive interference).`

### Re: Sounds Of Music C

Posted: September 5th, 2018, 6:03 pm
Hi there, here is my answer:
forgive my formatting
`dB = 10 log I / I (threshold intensity ) = 10 log  4(1 x 10^-4 W/m^2) / 1 x 10^-12 W/m^2 = 86dB, so final is 86dB`
Here is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straightened
out it would be 65 cm long. Using 343 m/s for the speed of sound, find the lowest
frequency that can be played, ignoring end corrections.

A wee bit word problem ### Re: Sounds Of Music C

Posted: September 5th, 2018, 6:22 pm
goblinrum wrote:Hi there, here is my answer:
forgive my formatting
`dB = 10 log I / I (threshold intensity ) = 10 log  4(1 x 10^-4 W/m^2) / 1 x 10^-12 W/m^2 = 86dB, so final is 86dB`
Here is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straightened
out it would be 65 cm long. Using 343 m/s for the speed of sound, find the lowest
frequency that can be played, ignoring end corrections.

A wee bit word problem So the fundamental frequency is equal to 2/1 Length. 65 cm * 2 = 130. 343 m/s = 1.3 m * frequency. After calculating, I got ~364 Hz.

### Re: Sounds Of Music C

Posted: September 5th, 2018, 6:23 pm
Darkmatter7 wrote:
goblinrum wrote:Hi there, here is my answer:
forgive my formatting
`dB = 10 log I / I (threshold intensity ) = 10 log  4(1 x 10^-4 W/m^2) / 1 x 10^-12 W/m^2 = 86dB, so final is 86dB`
Here is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straightened
out it would be 65 cm long. Using 343 m/s for the speed of sound, find the lowest
frequency that can be played, ignoring end corrections.

A wee bit word problem So the fundamental frequency is equal to 2/1 Length. 65 cm * 2 = 130. 343 m/s = 1.3 m * frequency. After calculating, I got ~364 Hz.
Sorry, forgot to tell you that you got the correct answer on my question ### Re: Sounds Of Music C

Posted: September 5th, 2018, 6:37 pm
Darkmatter7 wrote:
Darkmatter7 wrote:
goblinrum wrote:Hi there, here is my answer:
forgive my formatting
`dB = 10 log I / I (threshold intensity ) = 10 log  4(1 x 10^-4 W/m^2) / 1 x 10^-12 W/m^2 = 86dB, so final is 86dB`
Here is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straightened
out it would be 65 cm long. Using 343 m/s for the speed of sound, find the lowest
frequency that can be played, ignoring end corrections.

A wee bit word problem So the fundamental frequency is equal to 2/1 Length. 65 cm * 2 = 130. 343 m/s = 1.3 m * frequency. After calculating, I got ~364 Hz.
Sorry, forgot to tell you that you got the correct answer on my question Did you just do (343 m/s) / 1.3 m is 364Hz????

### Re: Sounds Of Music C

Posted: September 5th, 2018, 6:39 pm
Darkmatter7 wrote:
goblinrum wrote:Hi there, here is my answer:
forgive my formatting
`dB = 10 log I / I (threshold intensity ) = 10 log  4(1 x 10^-4 W/m^2) / 1 x 10^-12 W/m^2 = 86dB, so final is 86dB`
Here is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straightened
out it would be 65 cm long. Using 343 m/s for the speed of sound, find the lowest
frequency that can be played, ignoring end corrections.

A wee bit word problem So the fundamental frequency is equal to 2/1 Length. 65 cm * 2 = 130. 343 m/s = 1.3 m * frequency. After calculating, I got ~364 Hz.
I think you might be wrong or I might be wrong
Freq= speed/wavelength F= 343 /(2*.65meters)= 263.846 or 264 Hz.

### Re: Sounds Of Music C

Posted: September 5th, 2018, 6:39 pm
goblinrum wrote:
Darkmatter7 wrote:
Darkmatter7 wrote: So the fundamental frequency is equal to 2/1 Length. 65 cm * 2 = 130. 343 m/s = 1.3 m * frequency. After calculating, I got ~364 Hz.
Sorry, forgot to tell you that you got the correct answer on my question Did you just do (343 m/s) / 1.3 m is 364Hz????
264Hz. I miss typed -.-

### Re: Sounds Of Music C

Posted: September 6th, 2018, 8:20 am
A Solid has a bulk modulus of 2.3 × 10^7 N/m^2 and a density of 8.05 g/cm^3. What is the speed of sound in this solid?

### Re: Sounds Of Music C

Posted: September 6th, 2018, 8:48 am
Using the equation for speed of sound in a medium, V = sqrt(B/R), where B is the Bulk Modulus and R is rho, the density of the material.
Let's plug in the variables: sqrt((2.3*10^7)/(8.05*(10^3))) (converting density to metric units), which gives us 2857 m/s (if I didn't somehow mess up the calculation.