The amplitude determines the volume of the sound. This isn't affected by the motion of the sound source.
Hope this helps
-- The string is 1 m long. That's the radius of the circle that the mass is
traveling in. The circumference of the circle is (π) x (2R) = 2π meters .
-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .
-- Centripetal acceleration is V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²
-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =
64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .
</span>That's it. It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span>
</span>If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.
You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping.
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour ! This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.
Answer:
a) {[1.25 1.5 1.75 2.5 2.75]
[35 30 25 20 15] }
b) {[1.5 2 40]
[1.75 3 35]
[2.25 2 25]
[2.75 4 15]}
Explanation:
Matrix H: {[1.25 1.5 1.75 2 2.25 2.5 2.75]
[1 2 3 1 2 3 4]
[45 40 35 30 25 20 15]}
Its always important to get the dimensions of your matrix right. "Roman Columns" is the mental heuristic I use since a matrix is defined by its rows first and then its column such that a 2 X 5 matrix has 2 rows and 5 columns.
Next, it helps in the beginning to think of a matrix as a grid, labeling your rows with letters (A, B, C, ...) and your columns with numbers (1, 2, 3, ...).
For question a, we just want to take the elements A1, A2, A3, A6 and A7 from matrix H and make that the first row of matrix G. And then we will take the elements B3, B4, B5, B6 and B7 from matrix H as our second row in matrix G.
For question b, we will be taking columns from matrix H and making them rows in our matrix K. The second column of H looks like this:
{[1.5]
[2]
[40]}
Transposing this column will make our first row of K look like this:
{[1.5 2 40]}
Repeating for columns 3, 5 and 7 will give us the final matrix K as seen above.
