Sounds Of Music C

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Sounds Of Music C
Welcome to the 20182019 season! I'll start with some basics:
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?
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
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).
MIT ‘23
TJHSST ‘19
Longfellow MS
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TJHSST ‘19
Longfellow MS
See my user page for nationals medals and event supervising experience.
Re: Sounds Of Music C
Hi there, here is my answer:
forgive my formatting
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
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
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

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 Posts: 6
 Joined: June 29th, 2018, 9:51 am
Re: Sounds Of Music C
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.goblinrum wrote:Hi there, here is my answer:
forgive my formattingHere is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straighteneddB = 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
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

 Member
 Posts: 6
 Joined: June 29th, 2018, 9:51 am
Re: Sounds Of Music C
Sorry, forgot to tell you that you got the correct answer on my questionDarkmatter7 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.goblinrum wrote:Hi there, here is my answer:
forgive my formattingHere is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straighteneddB = 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
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
Did you just do (343 m/s) / 1.3 m is 364Hz????Darkmatter7 wrote:Sorry, forgot to tell you that you got the correct answer on my questionDarkmatter7 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.goblinrum wrote:Hi there, here is my answer:
forgive my formattingHere is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straighteneddB = 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
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

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 Joined: September 5th, 2018, 3:55 pm
 Division: C
 State: TX
Re: Sounds Of Music C
I think you might be wrong or I might be wrongDarkmatter7 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.goblinrum wrote:Hi there, here is my answer:
forgive my formattingHere is my question : A soprano saxophone is considered an open pipe resonator. If the pipe was straighteneddB = 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
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
Freq= speed/wavelength F= 343 /(2*.65meters)= 263.846 or 264 Hz.

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 Posts: 6
 Joined: June 29th, 2018, 9:51 am
Re: Sounds Of Music C
264Hz. I miss typed .goblinrum wrote:Did you just do (343 m/s) / 1.3 m is 364Hz????Darkmatter7 wrote:Sorry, forgot to tell you that you got the correct answer on my questionDarkmatter7 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.

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 Posts: 6
 Joined: June 29th, 2018, 9:51 am
Re: Sounds Of Music C
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?

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 Posts: 5
 Joined: June 17th, 2017, 3:16 pm
Re: Sounds Of Music C
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.
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.
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