Thanks for the responses. I could see somewhat of the usefulness in the development and design of digital musical instruments and perhaps even in designing recording studios. I actually have a friend in Dartmouth's music program and I've heard some good things about it and looking at this...
I am a physics and electrical engineering major and wish to know if there are any specific music career opportunities that utilize the concepts I've learned from both. I really have a passion for music production and any help would be greatly appreciated.
I used this method and set y = 0 for equations. I then solved for x and set both equations equal to each other as follows:
x^2+z^2=1/2
x^2+(z-1/2)^2=1/4
0=1/2-z^2
0=1/4-(z-1/2)^2
I got z to equal 1/2 and plugged it back into those 2 equations I got when I set y = 0 and got 1/2 for the...
Homework Statement
Find the volume outside the sphere x2 + y2 + z2 = 1/2 and inside the sphere x2 + y2 + z2 = z
2. The attempt at a solution
I've gotten as far as to visually seeing that's there's two spheres and determining that the radius of the first sphere is 1/√2. However, I'm...
I finally understand why (h +.25 m) is wrong. We had set the Equilibrium to be 0 when there was no compression. The change in height initially due to compression was (0-.25 m). However, in order to find the final height, it would simply be (0 + h). In setting the final height to (h + .25 m)...
So then the height, h, would just be that (h + .25 m) instead, since it opposes the initial height, when the spring was compressed and the height was negative.
My problem was understanding the whole "set your own GPE = 0 and stay consistent with it" sort of deal. My professor mentioned that...
Ah, should I have taken the direction into consideration then? If so, then it would just be (-.25 m - 0m) for the initial height and for the final height, (h - .25 m), where h represents the final height without the height compression (since it is asking how high will the mass go above the...
Thanks a lot for the insight! I always welcome and love explanations and I appreciate that you showed me this.
As for the part when the spring gets compressed and the mass is launched into the air, is it strange that I have GPE(i) and SPE(i) = GPE(f)?
When the spring is initially compressed...
The approach I wanted to take on was of static equilibrium and I think I understand what you mean now. If I was to then use Hooke's law and Solve for F and x would it just be:
F = m*a
= (150 kg) (9.8 m/s^2)
= 1470 N
x = .15 m
k = (1470 N) / (.15 m)
= 9800
Ah, thank you for the reply collinsmark. I'm still pretty stuck. I set the GPE = 0 when the mass was stretched down .15 m. Is there some sort of sign error? Should it be -mgh = .5kx^2?
Homework Statement
A 150 kg mass is attached to a vertical spring, stretching it by 15.0 cm. This same spring is fastened to the floor and compressed by 25.0 cm. A 5.00 kg mass is then placed on the spring. If the spring is released and the mass is launched straight up into the air by the...