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Body of mass 6 kg is performing SHM with amplitude 0.28 m and angular frequency 5.5 rad s"1 Calculate the displacement of the body, in m; to 2d.p, when the pot...

Question

Body of mass 6 kg is performing SHM with amplitude 0.28 m and angular frequency 5.5 rad s"1 Calculate the displacement of the body, in m; to 2d.p, when the potential energy of the system is equal to its kinetic energy:

body of mass 6 kg is performing SHM with amplitude 0.28 m and angular frequency 5.5 rad s"1 Calculate the displacement of the body, in m; to 2d.p, when the potential energy of the system is equal to its kinetic energy:



Answers

A body of mass $1.80 \mathrm{kg}$ executes SHM such that its displacement from equilibrium is given by $x=0.360 \cos (6.80 t),$ where $x$ is in metres and $t$ is in seconds. Determine: (a) the amplitude, frequency and period of the oscillations; (b) the total energy of the body; (c) the kinetic energy and the elastic potential energy of the body when the displacement is $0.125 \mathrm{m}$

As we all know that the period of mass attached to the spring is given by T is equal to two pi blue tender and by Kay, and the first constant K is equal to to buy Uh huh, Holy Square Multiplication m So simplifying it further than putting the value I can write the expression edge to buy multiplication 2.6 Holy square Multiplication 1.8 On solving I get 479.88 Newtons per meter. Now, as we all know, that the energy is equal to one by two k squared. So simplifying it, I can write one by two Multiplication 479.88 Newtons per meter multiplication 0.71 m squared, which is equal to 1.21 jewel as our answer.

In this problem. We all know that time period is equal to two pi route under em by K So the first constant K can be written it to buy f holy square multiplication m So just putting the value years So two pi multiplication 2.6 Holy square multiplication 1.8 and on solving a gate evaluate 479 0.88 Newton per meter Going forward as we all know that the value of energy is equal to half a k a square. This is the formula, so I can write. E is equal to one by two multiplication 479.88 Multiplication 0.0 71 is squared and on solving it. I get the value of e H 1.21 Jules, this is our end.

Mhm in this problem, we are going to calculate the speed of those lighting object at the point where the kinetic energy and potential in Nigeria are equal. So we need to calculate, uh, we since the kinetic energy and potential energy are equal at the human point So we can right here, potential energy equals two kinetic energy. This can be further written as the one you want by two. Cake square, which is the potential energies, equals to one divided by two m we square this, uh, one divided by two and we square is the kinetic energy. We can write the equation for this case spring constant as case equals two um, Omega Square here, this mega angular frequency of those lessons. And, uh, we can wait the speed as well as equals two omega squared of, uh, square minus X squared. Here, this is the amplitude. And this X is the displacement of those license. Using these two relations, we can write this equation as one divided by two mm. Omega square X square is equals to one divided by two mm into omega squared of a square minus omega, minus x square. So here we have X squared into whole square. Yeah. From here we can white the equation for this X as X equals two Eddie weighted by squared off to mhm. Now we can write the required number two, uh, this one, This equation number two as, uh, as a we as equals two 25 which is the angular frequency. And this f is the linear frequency squared of, uh, square minus, uh, s square, divided by two. So this can be simplified as a to buy F s square 25 a. And then here we have one divided by square root of two. So this can be further written as well as equals. Two squared of two. Bye f A. This will be a request number. Let's see three. Let's put the values into this question so it will be busy equals to square root of two into 3.14 into the frequency which is equals to zero point 377 herds into the value of which is equals two 28 point to multiply with Terrence part minus 2 m. So from here, we can ride the value for this we as well as equals two 0.47 Major. Persistent. So this is the resulted resulting to answer and of the question. Thank you.

So we have a mass equaling one point one five kilograms and then we have, ah, a function for motion core for the ex displacement. Since we'LL be accepting equaling point six five zero times co sign of eight point four zero t and we can relate this to the general equation of a Times Co sign of Omega T now for party. They're asking for the amplitude. This is simply going to be point six five zero meters because we're simply relating this general equation to the equation that they're giving us and then for being what they're asking for the frequency we know the the the angular velocity Omega equals two pi times f, which means that the frequencies and equal to omega divided by two pie So this will equal eight point four divided by two pie. And we're getting a frequency of one point three four of hurts now for part. See, they're asking us for the total amount of energy so eat total is going to be equaling one half em the max squared and well, plugging in for Visa Max, this is going to be equal to one half em time's omega, eh, squared and this is going to be able to one half times m. So one point one five time's omega. So eight point four times of point six five squared and regaining, eh Energy Total of seventeen point one for Jules. Now for Part D. They're asking us for the kinetic energy when X equals point three six zero meters. So the potential energy can be calculated as one half k x squared to the potential energy of a spring. This is going to be one half times the spring constant or rather, one half times M omega X squared. And this is going to be equal to one half times again. One point one five ah, eight point four times acts. So point three six squared. This is going to be equal to a five point two six jewels. So the kinetic energy is simply going to be equal to the total energy minus the potential energy. So this is going to be seventeen point one four minus five point two six. And we're getting eleven point eight eight jewels of kinetic energy when x equal ISS point three six zero meters. And that's the end of the solution. Thank you for watching


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