Sounds Of Music C

Post Reply
Darkmatter7
Member
Member
Posts: 6
Joined: June 29th, 2018, 9:51 am
Has thanked: 0
Been thanked: 0

Sounds Of Music C

Post by Darkmatter7 »

Welcome to the 2018-2019 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?
mjcox2000
Member
Member
Posts: 121
Joined: May 9th, 2014, 3:34 am
Division: Grad
State: VA
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by mjcox2000 »

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

See my user page for nationals medals and event supervising experience.
goblinrum
Member
Member
Posts: 3
Joined: June 29th, 2018, 6:32 am
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by goblinrum »

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 :)
Darkmatter7
Member
Member
Posts: 6
Joined: June 29th, 2018, 9:51 am
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by Darkmatter7 »

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.
Darkmatter7
Member
Member
Posts: 6
Joined: June 29th, 2018, 9:51 am
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by Darkmatter7 »

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 :D
goblinrum
Member
Member
Posts: 3
Joined: June 29th, 2018, 6:32 am
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by goblinrum »

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 :D
Did you just do (343 m/s) / 1.3 m is 364Hz????
HeadphoneMonster
Member
Member
Posts: 3
Joined: September 5th, 2018, 3:55 pm
Division: C
State: TX
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by HeadphoneMonster »

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.
Darkmatter7
Member
Member
Posts: 6
Joined: June 29th, 2018, 9:51 am
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by Darkmatter7 »

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 :D
Did you just do (343 m/s) / 1.3 m is 364Hz????
264Hz. I miss typed -.-
Darkmatter7
Member
Member
Posts: 6
Joined: June 29th, 2018, 9:51 am
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by Darkmatter7 »

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?
epicdragon44
Member
Member
Posts: 5
Joined: June 17th, 2017, 3:16 pm
Has thanked: 0
Been thanked: 0

Re: Sounds Of Music C

Post by epicdragon44 »

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.
Post Reply

Return to “2019 Question Marathons”

Who is online

Users browsing this forum: No registered users and 0 guests