Crave The Wave B

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UTF-8 U+6211 U+662F
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Re: Crave The Wave B

Postby UTF-8 U+6211 U+662F » May 2nd, 2015, 9:56 am

RontgensWallaby wrote:
UTF-8 U+6211 U+662F wrote:Oops, um, said that wrong. :oops:
A sound wave of 8 dB travels 1 meter and ends up being 5 dB. What is the volume of the attenuated sound wave after 1.5 meters?


Oh, ok.

ANS
1.6375 dB

Hmm, I got
Possible ANS
3.333 dB (By the inverse square law, 5 dB * 1 meter = x dB * 1.5 meters -> x = 3 1/3
Can you explain your work?

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Re: Crave The Wave B

Postby RontgensWallaby » May 2nd, 2015, 4:16 pm

Never mind. I used the cube of it. However, after considering my dumb mistake, I still got a different answer...

Ans
My final answer now is 2.78 dB. The problem can be set up like the exponential function ab^x, where a is the original value (8), b is the coefficient (5/8 or 0.625 since that is how much is lost after one meter), and x is the exponent (in this situation, the SQUARE of the distance because the surface area of a sphere is proportional to the square of its radius). Thus our equation is 8(0.625)^2.25, which turns out to round to 2.78 dB.


I think that is correct... Tell me if you agree.
Every great and deep difficulty bears in itself its own solution. It forces us to change our thinking in order to find it. - Niels Bohr

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Re: Crave The Wave B

Postby UTF-8 U+6211 U+662F » May 2nd, 2015, 4:23 pm

RontgensWallaby wrote:Never mind. I used the cube of it. However, after considering my dumb mistake, I still got a different answer...

Ans
My final answer now is 2.78 dB. The problem can be set up like the exponential function ab^x, where a is the original value (8), b is the coefficient (5/8 or 0.625 since that is how much is lost after one meter), and x is the exponent (in this situation, the SQUARE of the distance because the surface area of a sphere is proportional to the square of its radius). Thus our equation is 8(0.625)^2.25, which turns out to round to 2.78 dB.


I think that is correct... Tell me if you agree.

Decibels are already exponential, so it turns more into the inverse proportion law. So:

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Re: Crave The Wave B

Postby RontgensWallaby » May 2nd, 2015, 4:55 pm

UTF-8 U+6211 U+662F wrote:
RontgensWallaby wrote:Never mind. I used the cube of it. However, after considering my dumb mistake, I still got a different answer...

Ans
My final answer now is 2.78 dB. The problem can be set up like the exponential function ab^x, where a is the original value (8), b is the coefficient (5/8 or 0.625 since that is how much is lost after one meter), and x is the exponent (in this situation, the SQUARE of the distance because the surface area of a sphere is proportional to the square of its radius). Thus our equation is 8(0.625)^2.25, which turns out to round to 2.78 dB.


I think that is correct... Tell me if you agree.

Decibels are already exponential, so it turns more into the inverse proportion law. So:


OK, now I think I have it. Decibels are logarithmic, so I believe this is the way you'd do it...

Yet another possible explanation
A loudness ratio of 8 dB to 5 dB corresponds to an intensity ratio of 1 to 0.4762. Because intensity decreases with the square of the distance, the ratio of intensities between the intensity at the source to the intensity 1.5m away is 1 to (0.4762)^2.25 = 0.1884 or 5.307:1. This corresponds to a difference of 7.25 dB, so the resulting intensity is 0.75 dB.

I'm aware that my answer sounds really unrealistic but my calculations (I hope) are all correct and 8 dB is barely audible in the first place.
Every great and deep difficulty bears in itself its own solution. It forces us to change our thinking in order to find it. - Niels Bohr

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Re: Crave The Wave B

Postby UTF-8 U+6211 U+662F » May 2nd, 2015, 5:03 pm

RontgensWallaby wrote:
UTF-8 U+6211 U+662F wrote:
RontgensWallaby wrote:Never mind. I used the cube of it. However, after considering my dumb mistake, I still got a different answer...

Ans
My final answer now is 2.78 dB. The problem can be set up like the exponential function ab^x, where a is the original value (8), b is the coefficient (5/8 or 0.625 since that is how much is lost after one meter), and x is the exponent (in this situation, the SQUARE of the distance because the surface area of a sphere is proportional to the square of its radius). Thus our equation is 8(0.625)^2.25, which turns out to round to 2.78 dB.


I think that is correct... Tell me if you agree.

Decibels are already exponential, so it turns more into the inverse proportion law. So:


OK, now I think I have it. Decibels are logarithmic, so I believe this is the way you'd do it...

Yet another possible explanation
A loudness ratio of 8 dB to 5 dB corresponds to an intensity ratio of 1 to 0.4762. Because intensity decreases with the square of the distance, the ratio of intensities between the intensity at the source to the intensity 1.5m away is 1 to (0.4762)^2.25 = 0.1884 or 5.307:1. This corresponds to a difference of 7.25 dB, so the resulting intensity is 0.75 dB.

I'm aware that my answer sounds really unrealistic but my calculations (I hope) are all correct and 8 dB is barely audible in the first place.

Now, I'm confused. Somebody post a question.

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Re: Crave The Wave B

Postby RontgensWallaby » May 2nd, 2015, 5:13 pm

I'll make this one a bit easier.
What is the peak wavelength emitted by a blackbody with a temperature of 10000 K?
Every great and deep difficulty bears in itself its own solution. It forces us to change our thinking in order to find it. - Niels Bohr

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Re: Crave The Wave B

Postby RontgensWallaby » May 2nd, 2015, 5:16 pm

Just going to post the answer right here so that I don't forget it.

ANS
2.898*10^-7 m or 289.8 nm
Every great and deep difficulty bears in itself its own solution. It forces us to change our thinking in order to find it. - Niels Bohr

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Re: Crave The Wave B

Postby RontgensWallaby » May 8th, 2015, 3:30 pm

Found another good question, this time it's not original. Comes from Physics for Scientists and Engineers with Modern Physics.
Astronauts are visiting Planet X. They take a 2.5m string with a mass of 5g and a 1kg mass tied to one end. On the planet, the astronauts fix one end of the string, horizontally extend it for 2m, hold it as shown in this picture (image1.masterfile.com/em_w/03/65/18/632-03651827em.jpg) after 2m, and allow the remaining .5 m with the mass tied to it to hang freely. The astronauts then proceed to pluck the 2m long string, finding that the string forms a standing wave at 64 Hz and at 80 Hz but at no frequencies in between. What is the gravitational acceleration on Planet X?

A useful formula: v=root(T/d) where v is the speed of a wave on a string, T is the tension in millinewtons, and d is the linear density in grams per meter.
Every great and deep difficulty bears in itself its own solution. It forces us to change our thinking in order to find it. - Niels Bohr


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