A mass is oscillating with amplitude A at the end of a spring. Ho...

Question

A mass is oscillating with amplitude A at the end of a spring. How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

Accepted Answer

Hey everyone in this problem, we have an object of mass M that executes a simple harmonic motion when attached to a spring with spring constant K. The amplitude of the simple harmonic motion is A And we're asked to find the position of the object from the equilibrium position if the kinetic energy is double the potential energy. And we're told to express our answer in terms of the amplitude. A. Okay. Alright. So we're asked to find the position of the object and were given some information about the relationship between the kinetic energy and potential energy. So, let's think about mechanical energy here and let's recall, because we have no net external forces acting here, we're going to have mechanical energy conserved. What that means. Is that the mechanical energy at the point P that we're interested in this position that we're interested in is going to be equal to the mechanical energy at some other point in our system. Okay. And any other point in our system and we're gonna choose mechanical energy at the amplitude. A. Okay. And you're gonna see in a minute why we chose the amplitude A is going to simplify our calculations for us. Okay, Alright. So what makes up the mechanical energy? Well, it's going to be the kinetic energy. So we have the kinetic energy at point P Plus the elastic potential energy. We're going to call that you E. Okay. And in this case at point P. And same thing for the mechanical energy at the amplitude with the kinetic energy at the amplitude plus the elastic potential energy at the amplitude. All right now, when we're thinking about the kinetic energy at the amplitude. Ok, recall that the kinetic energy is one half M v squared. Ok. That the amplitude our object is going to be switching from going in one direction to going in the other direction. Okay? And it's gonna momentarily come to a stop. Okay? So the speed v at the amplitude is zero, which means that the kinetic energy at the amplitude is zero. Okay, so this term is going to go away entirely. Okay, so that's why we chose the point at the amplitude. Because this simplifies this calculation, this entire kinetic energy goes to zero. All right now, we're looking for the position of the object and the position is going to come into this elastic potential energy equation. Recall that elastic potential energy is one half K X squared. So we want to keep this and we're told that the kinetic energy is double the potential energy at this point. Okay, So what that means is K. P. Is going to be two times U E P. Okay, so if we replace that. And then on the left hand side we're gonna get three times the elastic potential energy at our point P is equal to the elastic potential energy at the amplitude. A Alright now, recall we just mentioned the equation for the elastic potential energy is one half K X squared. Okay, so on the left hand side we get three halves. Okay, XP this exposition that we're interested in squared. On the right hand side we have one half. Okay and the position at the amplitude is just gonna be a Yes we get one half K A squared. We want to solve for the position X. So we get that XP- 2ared is equal to the case. Will divide. The 1/2 will divide. Okay? And then we're gonna get 1/3 a squirt. Okay? And when we take the square root we get that the position X that we're interested in is going to be a lost or minus. Can we take the square root? We have the positive and negative route over the square root of three. Alright, so if we look at our answer choices we see is that we have answer choice. A. Okay, the position of the object from the equilibrium is going to be plus or minus. A over three to the exponents. One half. Okay and remember three to the exponent? One half. That's equivalent to the square root of three. Okay, so that is equivalent to the position we found. Thanks everyone for watching. I hope this video helped see you in the next one

A mass is oscillating with amplitude A at the end of a spring. How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

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