<p>I can see myself making the same mistake that you did on the U(r) thing. The time limit freaks me out and I have the tendency to rush through everything, making stupid mistakes along the way.</p>
<p>For the second one, its a second-order differential equation.
You’re not responsible for solving it at the Physics C level, but basically, you can guess a value for x (what kind of function can you differentiate twice and still get the original function?). Yes the answer is either sin or -cos (Simple Harmonic Motion oscillating functions).
What were the options? You could eliminate a lot like that.
I’d guess the answer would be something like -9cos(wt) or something…</p>
<p>Would it be a) 6pi ?
Period of a cosine curve is 2pi, so the period of the pendulum would be…</p>
<p>wait no, don’t think thats right.
What was the answer?</p>
<p>And btw, if you’re given the graphs of force as a function of position for objects undergoing SHM, how would you graph its potential energy function? Would it just be the antiderivatives of the given functions?</p>
<p>Are you sure it’s c? Think of the general simple harmonic oscillator displacement equation x=Acos(wt+phi). Differentiating that twice will give you a=-w^2(x) which is equal to -9x. Therefore w=3. And w=2pi/period so period=2pi/w and T=2pi/3. (answer D)</p>
<p>or…you could think that the net force on any SHO is -kx. So ma=-kx and a=-(k/m)x where (k/m) is then equal to 9. Period equals 2pi(sqrt(m/k)) and m/k is the reciprocal of 9. So T=2pi(sqrt(1/9))= 2pi/3. (again answer D)</p>
[quote=AxeBack] @feuxfollets: Its the same thing. An indefinite integral of dt actually has the limits of 0 to t (thats why it ends up becoming t by FTC).
<p>Hmmm yea I asked because on the answer key in my Barron’s book they just ignored the limits, but that was because they just threw everything into the C term… lol</p>
<p>Oh lol, I see what you mean.
Ok, for differential equations in Physics C, I don’t think you need to plug in limits.
Only when the question specifies you to solve for a certain time period, then I’d guess you need to do that (don’t think I’ve ever seen one).
So yeah, just separate variables and integrate as an indefinite integral.</p>
<p>Your constant will probably be taken care of with some given initial condition.</p>
<p>Consider a circuit consisting of the following elements in a single loop: A battery, a resistor, a switch and a fourth unknown element E. A spark is most likely to fly if:</p>
<p>A) E is a capacitor and the switch is being closed for the first time.
B) E is a capacitor and the switch is being opened after the circuit has reached equilibrium.
C) E is an inductor and the switch is being closed for the first time.
D) E is an inductor and the switch is being opened after the circuit has reached equilibrium.
E) E is a battery arranged in series with the other battery in the circuit.</p>
<p>@ Axeback: Answer is (D). Causes an open circuit containing an inductor. The inductor tries to resist change in current (right before opening, the current is maximum), but an open circuit causes the spark. Just curious, where did you find this question?</p>
<p>Oh yea…right. Forgot everything about inductors.
A friend asked me this while we were studying for the test on AIM lol.
I think he found it in the Barron’s book…not sure.
And this explanation is a lot better than his.</p>
<p>Wow I’m totally not ready for this. E/M is gonna be a *****</p>
<p>An ideal massless spring is fixed to a wall at one end. A block of mass M is at the other end, it oscillates with amplitude A on frictionless surface. The max. speed is v_m. Force constant=?</p>
<p>just use energy. you know that maximum PE occurs at distance A from the eq point, and this max PE = max KE which you have since you know the max velocity.</p>
<p>does anyone have like a quick review sheet explaining everything (especially electromagnetism) online somewhere? I am completely clueless about this stuff lol</p>
<p>another question:
how do you find potential difference between two points in a one loop circuit?
what formula do you use?
V = RI?
kirchoff’s rule for loops?</p>
<p>Potential difference? Well, depends on the resistance.</p>
<p>If there’s a resistor between the two points, then it’s IR… that’s the voltage drop.</p>
<p>Kirchoff’s loop rule states that for any loop, the potential difference = EMF, that is the total voltage drop around a loop must be equal to the emf.</p>
<p>To tomjonesistheman:
Say you have point x, and point y, and you wanna find the potential difference. Just go down the loop, taking any path from x to y. SO, you do Vx-whatever comes in the way of the path to y=Vy…thats all its simple</p>
<p>To the people discussing the second derivative = -9x</p>
<p>We know two things. The second derivative implies that a=-9x. Therefore F=ma=-9mx. DO you realize this equation? It’s proportional to Hooke’s law, where k = -9m, and x equals x. So, from period, we get T=2pi sq (m/k)… Plug in -9m for k and simplify, and you’ll get T=2pi/3</p>