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Homework Help: Simple pendulum with initial velocity

  1. Apr 12, 2014 #1

    I've attached a couple questions just because these are the type of questions I've had some difficulty solving since I don't recall addressing anything too similar too similar in the past (i.e., changing velocity and determining how the amplitude varies because of this). I know to go from x(t) to v(t) you take the derivative, and that x(t) = xmcos(ωt + ø) in general. Then v(t) = -ωxmsin(ωt + ø). When looking at questions similar to 39, when I see vo is now 4vo and asking for the amplitude change to be determined, I look at xm. This means the new amplitude would have to be 4xm since all other variables are being kept constant, no? Also, once again taking the derivative to find a(t), wouldn't the acceleration be 4 times as great once again?

    Is there any other fundamental expression relating amplitude with velocity or frequency or any other variable as in question 21 or in general? Once again, I am also slightly confused on how exactly question 25 was solved too. If there any good online sources I could be referred to that would be greatly appreciated! If I am just missing something here and if someone could point out what I am not looking at, that would be amazing too!


    Attached Files:

  2. jcsd
  3. Apr 12, 2014 #2


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    For the first question, I quite agree with you.

    You can view it from a conservation of energy perspective as well. When the mass is instantaneously at rest, all its KE has been converted to elastic PE in the spring, and this is also the point at which displacement = amplitude. At this point ##\frac{1}{2}kx^2 = \frac{1}{2}mv^2##, where ##v## is the initial velocity of the mass. So ##x \propto v##. The maximum acceleration also occurs at this point, and since ##F = -kx##, ##a \propto v##. Hence, I think D should be the right answer.

    As to no. 25, think about which trig ratios are positive and negative in which quadrants.
  4. Apr 12, 2014 #3


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    Yes and yes. The key must be wrong.

    You can also see this by considering the interplay of potential energy and kinetic energy. The max PE is proportional to xmax2 and the maximum KE is proportional to vmax2. In these cases vmax is v0 or 4v0.

    I don't see problem 21.

    You might get better response if you would include the images in your post.

    By The Way: Don't ask unrelated questions in the same post. (In one you ask for some specific help. In the other, some more general help.)

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