26 thoughts on “Homework #12”

  1. slashley

    Hello. I’ve been trying to find the Part A of problem 14.1 on the online homework… I have the amplitude and the frequency. Don’t I just plug those into:
    -(2 pi f) A sin(2 pi f t)
    …. to get the velocity? I’ve been trying that [& I did convert cm to m] but it still doesn’t give me a correct answer.. and for t I’m using the t that was given [ie: Determine the velocity at t = .4 seconds ]

  3. mtheriau

    For the written homework, I am having trouble finding the equilibrium position in order to calculate the amplitude. The 10 cm and 60 cm would be the min and max right?

  4. physics131-willocq Post author

    The minimum position is 10 cm and the maximum position is 60 cm. These are the positions for extreme displacements away from the equilibrium position. You can assume the equilibrium position is halfway between those two positions.

  5. slashley

    For the online homework, problem 14.16, I’m unable to find Part B. I used the x they gave me for Amplitude and tried using that in the equation I used for Part A [just solving for speed instead, though] but it didn’t work… can you guide me along with this problem? Thank you

  8. physics131-willocq Post author

    Use conservation of energy to find the amplitude. You can calculate the initial kinetic energy and equate that to the spring potential energy for the maximum displacement (which then involves the amplitude).

  9. banoel

    For online hw 14.1 part b i keep getting the same answer for k, and when I check it into the frequency equation it works.

    I converted the mass from g to kg, and reworked the frequency equation to solve for the spring constant k. I don’t understand why my answer is incorrect. Any help would be immensely appreciated. Thank you.

  10. rdetroy

    I can seem to figure out how to calculte the frequency for part B of 14.1. I was wondering if you have any hints on how I would obtain this quantity.

  12. physics131-willocq Post author

    For 14.1 part B, you need to look at the specific expression for x(t), more exactly what the argument of the cosine function is. The form of the primary equation for position is x(t) = A cos(2pi f t), so if you are given x(t) = (50 m) cos(60 t), then we know that 2pi f = 60 Hz.
    From that you can get the frequency f.

  13. rdetroy

    So I am still confused on how to obtain the amplitude for 14.16… From your post, it seems that you use conservation of energy however, we do not know x either, so how do we find A? If you have any suggestions please let me know. Thank you!

  14. physics131-willocq Post author

    For 14.16 we do know that the mechanical energy is conserved so the maximum kinetic energy = maximum spring potential energy and we can calculate the maximum kinetic energy from the information that is given.

  15. gsakalin

    I am confused about 14.16 as well. I think the question is misleading. It asks “What is the block’s speed at the point where x= 0.30 A?”, does that mean what is its speed at 0.30 * A (which we got in part a) or is it just a value for x to put in the equation we used in class. I tried it both ways but I cant seem to get the right answer.

  19. rdetroy

    once we find x, how do we solve for the velocity? I tried using conservation of energy, but I am not getting the right answer…just let me know when you get the chance.

  20. physics131-willocq Post author

    You do know the maximum kinetic energy since you are given the initial velocity, this is for x=0 since the mass is at its equilibrium position. You then can write that the mechanical energy at x=0 must be the same as that at x=0.30 A (or wherever you want to know it) and this energy involves both kinetic and spring potential energies (and thus the velocity of the block at that position).

  22. physics131-willocq Post author

    For the gravity on another planet problem, you should refer to the equation giving the period of oscillation of a pendulum as a function of the acceleration due to gravity.

  23. rdetroy

    I actually have two quetsions regarding the online practice exam. Question #1: I am stuck on part b of 10.66 and I was wondering if you had any suggestions on how to approach the problem. Question #2: for problem 11.28, how do we figure out the amount of energy that needs to be extracted from the hot resevoir to produce 1000 J? Just let me know when you get the chance:)

  24. rdetroy

    so I have one more question….for part c of question 11.42, how do we calculate the force?

  25. physics131-willocq Post author

    For problem 10.66 part B, you need to treat the collision between the two boxes as an elastic collision. The final velocity of the block initially sliding down can be calculated using the equation I showed in my lecture notes in the case of perfectly elastic collisions. Once you have the velocity after the rebound, you can calculate the final height using energy conservation.

    For problem 11.28, you need to compute the maximum efficiency for the heat engine. Then calculate only a fraction of that as being the actual efficiency you can extract energy from the heat engine. The given 1000 J energy is what you get, you need to figure out what you pay.

    For problem 11.42 part C, you need to calculate the force knowing the work done and distance over which the force is applied. Since W = F d, then we have F = W/d.

    HTH.

  26. rdetroy

    for 10.66, I still don’t get how we can obtain the final velocity for package one. If both of the packages start from rest than the intial velocity=0 and then we are left with 0 = m(vf)+2m(vf), with two unknowns. I am not sure if I am just approaching the problem wrong if there is something else I am missing.

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