UTF-8 U+6211 U+662F wrote:Oops, um, said that wrong.
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.
1.6375 dB
Hmm, I got
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?
Re: Crave The Wave B
Posted: May 2nd, 2015, 4:16 pm
by RontgensWallaby
Never mind. I used the cube of it. However, after considering my dumb mistake, I still got a different answer...
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.
Re: Crave The Wave B
Posted: May 2nd, 2015, 4:23 pm
by UTF-8 U+6211 U+662F
RontgensWallaby wrote:Never mind. I used the cube of it. However, after considering my dumb mistake, I still got a different answer...
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:
Re: Crave The Wave B
Posted: May 2nd, 2015, 4:55 pm
by RontgensWallaby
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...
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...
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.
Re: Crave The Wave B
Posted: May 2nd, 2015, 5:03 pm
by UTF-8 U+6211 U+662F
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...
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...
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.
Re: Crave The Wave B
Posted: May 2nd, 2015, 5:13 pm
by RontgensWallaby
I'll make this one a bit easier.
What is the peak wavelength emitted by a blackbody with a temperature of 10000 K?
Re: Crave The Wave B
Posted: May 2nd, 2015, 5:16 pm
by RontgensWallaby
Just going to post the answer right here so that I don't forget it.
2.898*10^-7 m or 289.8 nm
Re: Crave The Wave B
Posted: May 8th, 2015, 3:30 pm
by RontgensWallaby
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.
