(I) A 1.28-kg mass oscillates according to the equation 𝓍 = 0...

Question

(I) A 1.28-kg mass oscillates according to the equation 𝓍 = 0.650 cos7.40 t where 𝓍 is in meters and t in seconds. Determine the kinetic energy and potential energy when 𝓍 = 0.260 m.

Accepted Answer

Welcome back. Everyone. In this problem. A metal ball of 2.43 kg is attached to the free end of an elastic spring. The other end of which is fixed, the ball is pulled from the equilibrium and released which causes it to oscillate. The displacement of the ball from the equilibrium is given by X equals 0.43 costs 5.3 T where the quantities are measured in the si units determine the kinetic and potential energy of the metal ball corresponding to the displacement X equals 0.215 m. For our answer choices A says that the kinetic energy is 0.893 Joules while the potential energy is 0.298 Joules B says it's 6.01 Jews and 0.298 Jews respectively. C 4.73 Jews and 6.31 Jews respectively. And D 4.73 Jews and 1.58 Jews respectively. Now let's first make a note of all the information that we have here because remember we're trying to find the kinetic energy and the potential energy of our mental bar. No, so far. OK. We're given the mass of the ball to be equal to 2.43 kg. OK. Next, we have an equation for the position from equilibrium as X equals 0.435 0.3 T. But recall that our equilibrium position is generally written in the form X equals A ca omega T where X is our amplitude and omega is our angular frequency. So if we compare the general form to the equation that we have, this tells us then that our amplitude A is 0.43 m and our angular frequency omega is 5.3 radiance per second. OK. Also we know that the final position X is equal to 0.215 m. OK. So rather, I should say here not recall but from that expression and no, we want to use somehow use all of this information to figure out the kinetic and potential energies. Now, what do we know? Well here, this represents simple harmonic motion. And for simple harmonic motion, the total mechanical energy that is the kinetic and potential energy remains constant if there is no energy dissipation. So that means then the total energy OK. The total energy is going to be equal to our potential energy U plus our kinetic energy K. Now let's ask ourselves, what do we know about our kinetic and potential energies? Well, we know that for our potential energy, it's equal to a half of kx squared. OK. Where K is equal to M omega squared, which tells us then that we can either take it a step further to say that our potential energy U is going to be equal to a half of M omega squared multiplied by X squared. Now, with that in mind, we can use this to figure out our potential energy because this means then that our potential energy U is going to be equal to a half of our mass, which is let me get rid of this here, which is 2.43 kg. OK. Multiplied by our angular frequency omega 5.3 radiance per second squared multiplied by our value of X which is 0.215 m of final position squared. Now, when we substitute, OK, we get our potential energy to be equal to 1.5776 jules. Now this is good. We've solved one part of the problem already. Now we just need to figure out our kinetic energy. OK. So what do we know about our kinetic energy? Well, OK. We know we may not have something for kinetic energy here, but we know that our total energy E OK is going to be equal to a half of K A squared where we already defined K as M omega squared. So that's actually equal to a half of M omega squared A squared. Now, while we don't have K, we can figure K out because that means this total energy is going to be equal to U plus K. So that tells us then that K is going to be equal to that value E minus our potential energy. So it's gonna be a half of M omega squared A squared minus 1.5776 Jules. Again, let's see if we can substitute our values to solve. This node is going to be a half of 2.43 kg, multiplied by 5.3 radiance per second squared. OK. Multiplied by or amplitude, 0.43 m squared minus 1.5776 Jules. And now we can put this expression into our calculator. OK? When we do that, we get the kinetic energy to be equal to 4.7329 Jules. So our kinetic energy is 4.7329 joules and our potential energy is 1.5776 joules. Notice all of our answers were into three significant figures. So if we do the same, we get the potential energy to be equal to 1.58 joules. OK? And we get the kinetic energy to be equal to 4.73 joules. If we go back to our answer choices, OK. That means that tells us then that D is the correct answer. Thanks a lot for watching everyone. I hope this video helped

(I) A 1.28-kg mass oscillates according to the equation 𝓍 = 0.650 cos7.40 t where 𝓍 is in meters and t in seconds. Determine the kinetic energy and potential energy when 𝓍 = 0.260 m.

Key Concepts

Simple Harmonic Motion (SHM)

Kinetic Energy (KE)

Potential Energy (PE)

Watch next