Crave The Wave B

[UTF-8 U+6211 U+662F]

In retrospect, the question is unsolvable... Try it but replace the 6s with just 1s and see how what you get compares to mine.

That's why I didn't reply.

[RontgensWallaby]

Anyone going to solve the problem where 6 seconds is replaced by 1 second?

[RontgensWallaby]

Anyone going to solve the problem where 6 seconds is replaced by 1 second?

As no one seems to be answering the question, I'll just put that the answer is 393.474 m/s.

[UTF-8 U+6211 U+662F]

Anyone going to solve the problem where 6 seconds is replaced by 1 second?

As no one seems to be answering the question, I'll just put that the answer is 393.474 m/s.

???

[RontgensWallaby]

Here's how you'd do it. Because the observer hears the sound after the plane passes by him, the plane has already broken the sound barrier and thus the sound is propagating backwards from it in the shape of a cone. (If you don't understand this there are plenty of online tutorials). For this problem, we'll make a simple diagram of when the observer first hears the sound. A cross-section of the sound cone is a triangle, which for simplicity's sake we will divide into a right triangle, as we can ignore the top half of the sound cone. The link below shows a picture of the right triangle we will use, where point A is the location of the observer, point B is the location of the plane, side a is the path taken by the plane, and side c is the edge of the sound cone (remember, this is the exact point where the sound reaches the observer). Because the plane constantly travels 700m above the ground, the length of side b is 700m. And since the observer hears the sound 1s after the plane passes directly overhead, side a is 1s multiplied by the velocity, which we'll refer to as x; therefore, side a is x. Now, angle B is what's called the Mach angle, and it has a special property. The sine of the Mach angle is equal to the speed of sound divided by the velocity of the plane. Therefore, we can create a similar right triangle (with the same Mach angle); however, in this triangle, side b is 343 (the speed of sound) and side c is x (the velocity of the plane). Because the side length ratio of our first triangle to our second triangle is 700/343, or about 2.041, side c in our first triangle is equal to 700x/343. Applying the Pythagorean Theorem to this first triangle, who has side a as x, side b as 700m, and side c as 700x/343, we can find that x=393.474m/s.





http://www.montereyinstitute.org/course ... age107.gif
Ignore the 30 deg mark at angle B, and the side labeled "4" will be referred to as side "b".

[RontgensWallaby]

OK. Someone else please post another question

[UTF-8 U+6211 U+662F]

Here's how you'd do it. Because the observer hears the sound after the plane passes by him, the plane has already broken the sound barrier and thus the sound is propagating backwards from it in the shape of a cone. (If you don't understand this there are plenty of online tutorials). For this problem, we'll make a simple diagram of when the observer first hears the sound. A cross-section of the sound cone is a triangle, which for simplicity's sake we will divide into a right triangle, as we can ignore the top half of the sound cone. The link below shows a picture of the right triangle we will use, where point A is the location of the observer, point B is the location of the plane, side a is the path taken by the plane, and side c is the edge of the sound cone (remember, this is the exact point where the sound reaches the observer). Because the plane constantly travels 700m above the ground, the length of side b is 700m. And since the observer hears the sound 1s after the plane passes directly overhead, side a is 1s multiplied by the velocity, which we'll refer to as x; therefore, side a is x. Now, angle B is what's called the Mach angle, and it has a special property. The sine of the Mach angle is equal to the speed of sound divided by the velocity of the plane. Therefore, we can create a similar right triangle (with the same Mach angle); however, in this triangle, side b is 343 (the speed of sound) and side c is x (the velocity of the plane). Because the side length ratio of our first triangle to our second triangle is 700/343, or about 2.041, side c in our first triangle is equal to 700x/343. Applying the Pythagorean Theorem to this first triangle, who has side a as x, side b as 700m, and side c as 700x/343, we can find that x=393.474m/s.





http://www.montereyinstitute.org/course ... age107.gif
Ignore the 30 deg mark at angle B, and the side labeled "4" will be referred to as side "b".

Okay. I'll ask a question:
A sound wave of 8 dB travels 1 m. What is the volume of the attenuated sound wave?

[RontgensWallaby]

Again, not sure if this is right.
If it's wrong, I'd appreciate if you sent me a link about it.

ANS (Click to Show Hidden Text)

7.385 dB

[UTF-8 U+6211 U+662F]

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?

[RontgensWallaby]

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.

ANS (Click to Show Hidden Text)

1.6375 dB