## Sounds Of Music C

epicdragon44
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### Re: Sounds Of Music C

Alright, since nothing seems to be happening let me give this thread a kick.

Question: This is a monochord:

It is a instrument composed of a single resonating box and a single string. Different notes can be played by adjusting the tension in the string.

There are two monochords tuning against each other. Both have strings of 0.14 g/m linear mass density and 0.5 m length. One of them (Let's call it Instrument A) is tuned so that the tension in the string is exactly 100 N. The other's (Instrument B's) tension is unknown. Given that the beat frequency heard when both instruments are played is 15 Hz, what are the two possible tensions, in Newtons (N), of Instrument B?

Solution (assuming I didn't calculate anything wrong, which I might have):Known information: Both strings have a linear mass density of 0.00014 kg/m (don't forget to convert the units!) and a wavelength of 2*0.5 (if you don't know where I got the 2, go back to Khanacademy) = 1 m. Instrument A has tension of 100 N, and we're trying to find the tension of Instrument B.
There are two key equations here:
V = λf
V = sqrt(Ft / (m/L)) - Note that Ft is one variable: it's the force of tension. m/L is also one variable, in this case: its our linear mass density.

If we set the equations equal to each other, we get the most important single 21 characters of your life xD
λf = sqrt(Ft / (m/L))
If we plug in the numbers we know, we can find the frequency of Instrument A:
1 * f = sqrt(100/0.00014)
f = 845.15 for Instrument A.

Now that we know the frequency, we can simply add and subtract our beat frequency to get the two possible frequencies of Instrument B (once again, if you don't know why, go back to studying. Giancoli in this case is great)
The two possible frequencies are 845.15+15 = 860.15 Hz, and 845.15-15 = 830.15 Hz. Let's tackle them one at a time...

Working backwards using the same 21-character equation, 860.15 * 1 = sqrt(Ft / 0.00014)
Ft = 103.58 N

Working backwards the same way with the other possible frequency, 830.15 * 1 = sqrt(Ft / 0.00014)
Ft = 96.48 N

To double check our answer for reasonable-ness, just see that Instrument A's tension was 100, and the beat frequency is pretty small so our deviation from that shouldn't be too large as well. As you can tell, it isn't.

Darkmatter7
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### Re: Sounds Of Music C

epicdragon44 wrote:Alright, since nothing seems to be happening let me give this thread a kick.

Question: This is a monochord:

It is a instrument composed of a single resonating box and a single string. Different notes can be played by adjusting the tension in the string.

There are two monochords tuning against each other. Both have strings of 0.14 g/m linear mass density and 0.5 m length. One of them (Let's call it Instrument A) is tuned so that the tension in the string is exactly 100 N. The other's (Instrument B's) tension is unknown. Given that the beat frequency heard when both instruments are played is 15 Hz, what are the two possible tensions, in Newtons (N), of Instrument B?

Solution (assuming I didn't calculate anything wrong, which I might have):Known information: Both strings have a linear mass density of 0.00014 kg/m (don't forget to convert the units!) and a wavelength of 2*0.5 (if you don't know where I got the 2, go back to Khanacademy) = 1 m. Instrument A has tension of 100 N, and we're trying to find the tension of Instrument B.
There are two key equations here:
V = λf
V = sqrt(Ft / (m/L)) - Note that Ft is one variable: it's the force of tension. m/L is also one variable, in this case: its our linear mass density.

If we set the equations equal to each other, we get the most important single 21 characters of your life xD
λf = sqrt(Ft / (m/L))
If we plug in the numbers we know, we can find the frequency of Instrument A:
1 * f = sqrt(100/0.00014)
f = 845.15 for Instrument A.

Now that we know the frequency, we can simply add and subtract our beat frequency to get the two possible frequencies of Instrument B (once again, if you don't know why, go back to studying. Giancoli in this case is great)
The two possible frequencies are 845.15+15 = 860.15 Hz, and 845.15-15 = 830.15 Hz. Let's tackle them one at a time...

Working backwards using the same 21-character equation, 860.15 * 1 = sqrt(Ft / 0.00014)
Ft = 103.58 N

Working backwards the same way with the other possible frequency, 830.15 * 1 = sqrt(Ft / 0.00014)
Ft = 96.48 N

To double check our answer for reasonable-ness, just see that Instrument A's tension was 100, and the beat frequency is pretty small so our deviation from that shouldn't be too large as well. As you can tell, it isn't.

Ok so here's my answer: T = 4f^2Lm
This is the equation for the tension of a standing string
T = 100N
L = .5 m
m = .14g/m * .5m
(Dont forget to convert m to kg)
So after solving for frequency, you get 845.15 HZ. Since the beat frequency is 15, by adding or substracting 15, you can get the two possible values for the frequency of string B. After solving for the Tension using the equation above (substituing the new frequencies):
T(b) = 103.58N
T(b) = 96.48N
Alright since I got that correct its my turn!

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### Re: Sounds Of Music C

Since no one has posted a question:

An empty 20 oz plastic soda bottle acts as a Helmholtz resonator. If the bottle's neck has a cross-sectional area of 2 cm^2 and is 1 cm long, and ambient conditions cause the speed of sound to be 340m/s, what is the bottle's resonant frequency?
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### Re: Sounds Of Music C

mjcox2000 wrote:Since no one has posted a question:

An empty 20 oz plastic soda bottle acts as a Helmholtz resonator. If the bottle's neck has a cross-sectional area of 2 cm^2 and is 1 cm long, and ambient conditions cause the speed of sound to be 340m/s, what is the bottle's resonant frequency?
$\frac{340 m/s}{2\pi}\cdot\sqrt{\frac{2 cm^2}{450 cm^3 \cdot 1 cm}}\cdot\frac{100 cm}{1 m}=361 Hz$, where 450 mL is the volume of a 20 oz soda bottle (obtained through DuckDuckGo)
Someone else can go because I'm clueless.

mjcox2000
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### Re: Sounds Of Music C

UTF-8 U+6211 U+662F wrote:
mjcox2000 wrote:Since no one has posted a question:

An empty 20 oz plastic soda bottle acts as a Helmholtz resonator. If the bottle's neck has a cross-sectional area of 2 cm^2 and is 1 cm long, and ambient conditions cause the speed of sound to be 340m/s, what is the bottle's resonant frequency?
$\frac{340 m/s}{2\pi}\cdot\sqrt{\frac{2 cm^2}{450 cm^3 \cdot 1 cm}}\cdot\frac{100 cm}{1 m}=361 Hz$, where 450 mL is the volume of a 20 oz soda bottle (obtained through DuckDuckGo)
Someone else can go because I'm clueless.
A 20oz bottle is actually 591mL, not 450mL, which makes the answer 315Hz, but other than that the math is right.
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### Re: Sounds Of Music C

mjcox2000 wrote:
UTF-8 U+6211 U+662F wrote:
mjcox2000 wrote:Since no one has posted a question:

An empty 20 oz plastic soda bottle acts as a Helmholtz resonator. If the bottle's neck has a cross-sectional area of 2 cm^2 and is 1 cm long, and ambient conditions cause the speed of sound to be 340m/s, what is the bottle's resonant frequency?
$\frac{340 m/s}{2\pi}\cdot\sqrt{\frac{2 cm^2}{450 cm^3 \cdot 1 cm}}\cdot\frac{100 cm}{1 m}=361 Hz$, where 450 mL is the volume of a 20 oz soda bottle (obtained through DuckDuckGo)
Someone else can go because I'm clueless.
A 20oz bottle is actually 591mL, not 450mL, which makes the answer 315Hz, but other than that the math is right.

Explain the relationship between intensity and loudness. What is the decibel scale?
Last edited by UTF-8 U+6211 U+662F on September 26th, 2018, 4:39 pm, edited 1 time in total.

goblinrum
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### Re: Sounds Of Music C

UTF-8 U+6211 U+662F wrote:
mjcox2000 wrote:
UTF-8 U+6211 U+662F wrote:
$\frac{340 m/s}{2\pi}\cdot\sqrt{\frac{2 cm^2}{450 cm^3 \cdot 1 cm}}\cdot\frac{100 cm}{1 m}=361 Hz$, where 450 mL is the volume of a 20 oz soda bottle (obtained through DuckDuckGo)
Someone else can go because I'm clueless.
A 20oz bottle is actually 591mL, not 450mL, which makes the answer 315Hz, but other than that the math is right.

Explain relationship between intensity and loudness. What is the decibel scale?
[spoiler]Intensity is a measure of the energy transmitted by the wave in watts per square meter, while loudness is the perceived loud/soft identification based of a reference intensity. The decibel scale is a logarithmic scale based on the threshold of hearing at 10^-12 W/m^2 to provide a measurement of loudness. Every 10 dB increase corresponds to 10 times louder.[/spoiler]

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### Re: Sounds Of Music C

goblinrum wrote:
UTF-8 U+6211 U+662F wrote:
mjcox2000 wrote: A 20oz bottle is actually 591mL, not 450mL, which makes the answer 315Hz, but other than that the math is right.

Explain relationship between intensity and loudness. What is the decibel scale?
[spoiler]Intensity is a measure of the energy transmitted by the wave in watts per square meter, while loudness is the perceived loud/soft identification based of a reference intensity. The decibel scale is a logarithmic scale based on the threshold of hearing at 10^-12 W/m^2 to provide a measurement of loudness. Every 10 dB increase corresponds to 10 times louder.[/spoiler]
Yep, your turn, although you don't need to spoiler the text if you're putting it in an answer box

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### Re: Sounds Of Music C

goblinrum wrote:
UTF-8 U+6211 U+662F wrote:
mjcox2000 wrote: A 20oz bottle is actually 591mL, not 450mL, which makes the answer 315Hz, but other than that the math is right.

Explain relationship between intensity and loudness. What is the decibel scale?
[spoiler]Intensity is a measure of the energy transmitted by the wave in watts per square meter, while loudness is the perceived loud/soft identification based of a reference intensity. The decibel scale is a logarithmic scale based on the threshold of hearing at 10^-12 W/m^2 to provide a measurement of loudness. Every 10 dB increase corresponds to 10 times louder.[/spoiler]
One minor point: while a 10dB increase does indeed correspond to 10 times the sound intensity, humans perceive it to be just twice as loud. That is, after a 10dB increase, we'd say that the intensity is multiplied by 10, but the loudness is doubled.
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epicdragon44
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### Re: Sounds Of Music C

[img]http://latex.codecogs.com/png.latex?\fr ... }=361%20Hz[/img]
Regarding this equation, could anyone link me to the original equation! That would be much appreciated.

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