Recent content by Djhar

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...

I appreciate the help but I need to strictly use integrals, even if that involves triple integrals

Would that then involve the subtraction of the volume of the first sphere from the second sphere?

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...