Calculate the ratio of the kinetic energy to the potential energy of a simple harmonic oscillator when its displacement is half its amplitude.
KE=1/2mv2 = 1/2kA2sin2(wt)
What I'm thinking is that when the displacement is half the amplitude, the simple harmonic oscillator is halfway to the equilibrium point. When the displacement is half of the amplitude, is the potential energy half of its maximum value? At this point, wouldn't the kinetic energy be half of its maximum value as well? I assumed this is true and attempted to answer the question, which yielded an answer of 1, which is incorrect. Is there something wrong with my logic? Answers and Replies
Matterwave
Yes, since the potential energy term goes as the square of the displacement, and not linearly with displacement, half the amplitude does not correspond to half the maximum potential energy.
Just plug x=1/2A into your equation for U=1/2kx^2 and then use conservation of energy to find the kinetic energy at that point.
Oh, I see what you're saying...but I'm a little lost at the conservation of energy part. I don't understand how to use that here. I know the potential energy is equal to 1/2k(A/2)^2 but isn't the kinetic energy just equal to 1/2mv^2? I understand that kinetic energy is related to the amplitude of the motion because velocity is related to amplitude...but I'm still not able to see how they are connected.
Matterwave
The total energy of the system E=T+U where T is kinetic, and U is potential energy. If you have U and E you can find T. In a SHO E is a constant and does not change. You just found U. Do you know what E is?
I took a closer look at my notes. Is the total energy equal to the square of the amplitude?
Oh woops, I don't think that makes sense.
Matterwave
Ok, at maximum amplitude, v=0, correct? That means at maximum amplitude, T=0. Now if E=T+U (at ALL times), and if T=0 then E=?
Yup, that makes sense...so E = U = 1/2kA^2, so at this point U is at its maximum value. As you said, for a simple harmonic oscillator, E is a constant...so this would mean that E is always equal to 1/2kA^2, right? E=1/2kA^2
U=1/8kA^2
T=?
Is the above correct? Now I'm starting to understand what you're saying. We did some problems like this in class...if you don't know E, you can always choose the point where either T or U is equal to 0 to figure it out...right? I mean, you'd only have to do that if you didn't know one or the either, like in this question.
Matterwave
Yes, you are on the right track. Now just figure out T from the above, and take a ratio.
The important thing to note here is that in a SHO, there is no damping. As such there is no dissipation of the energy, so E remains constant.
I got T=3, which is the correct answer. Thanks so much for all your help! :)
Answer  Hint: In physics, SHM stands for Simple Harmonic Motion which is defined as the motion of a body between two fixed points and it vibrates from one point to another and came back to its original point and this motion repeats several times with specific time period is called Simple Harmonic Motion. Complete step by step answer: (b) In SHM, the total energy of a body is defined as, $E = \dfrac{1}{2}k{x_m}^2$ where $k$ is called force constant. Potential energy is given as $U = \dfrac{1}{2}k\dfrac{{{x_m}^2}}{4}$ since it’s given that amplitude is half of maximum amplitude. $U = k\dfrac{{{x_m}^2}}{8}$Now, taking the ratios of Potential energy and total energy we get,$\dfrac{U}{E} = \dfrac{2}{8}$$\therefore U = 0.25E$Hence the fraction of potential energy to total energy is $0.25$. (a) Now, as we know, $E = T + U$ the, we can write it as:$\dfrac{T}{E} = 1 - \dfrac{U}{E}$From part (b) we know that $\dfrac{U}{E} = \dfrac{2}{8}$ put this value in above equation we get,$\dfrac{T}{E} = 1 - 0.25$$\therefore T = 0.75E$Hence, the fraction of Kinetic energy to total energy is $0.75$. (c) As, we know total energy of system is $E = \dfrac{1}{4}k{x_m}^2$ and let $x$ be the displacement at which potential energy became twice of total energy and we write it as: $U = \dfrac{1}{2}k{x^2}$ Now, both energies are equal in this displacement hence, so$\dfrac{{{x_m}^2}}{4} = \dfrac{{{x^2}}}{2}$$\therefore x = \dfrac{{{x_m}}}{{\sqrt 2 }}$Hence, the displacement at which total energy became half of potential energy is $x = \dfrac{{{x_m}}}{{\sqrt 2 }}$. Note:Remember, Using the law of conservation of energy we know, at any point of the motion in simple harmonic motion, the total energy at a point is always equals to the sum of kinetic energy and potential energy of the body at that point and it’s written as $E = T + U$. |
When the displacement is half the amplitude in a SHM The ratio of potential energy to the total energy is?
Pos Terkait
Copyright © 2024 membukakan Inc.
