A 0.500-kg glider, attached to the end of an ideal spring with fo...

Question

A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (e) the total mechanical energy of the glider at any point in its motion

Accepted Answer

Welcome back everybody. We are taking a look at some object on a flat surface connected to a spring. Now it is going back and forth following simple harmonic motion here. And we're told a couple different things, you're told that the mass of the object is 225 g or 2250.225 kg. We are told that there is a spring constant of 100 and 25 newtons per meter. We are also told that the maximum displacement Is 0.08 m or eight cm. And we are tasked with finding what the total mechanical energy is of the system. Well, according to the conservation of mechanical engineering energy, sorry, we know that this is just equal to the kinetic energy plus the potential energy of the system. Now, following simple harmonic motion, we actually have formulas for each of these. They're a little complex, but we'll be able to simplify simplify it down a lot. But let me just write them out first for kinetic energy, we have one half times the mass times negative, maximum displacement times omega times the sine of omega times time plus by all squared plus for our potential energy, it's very, very similar. We have one half pay as the displacement. Oh, sign of omega T Plus by Weird. Now let's simplify some things here. First, I'm going to distribute this little exponent here. Right? So we get that our total mechanical energy is equal to one half m times omega squared times a squared times this sine squared of omega T plus five Plus 1/2 pay a squared oh sine squared omega T plus by now our spring constant you know is just equal to the mass times omega squared. I'm actually gonna pull out this term and this term leaving us with this one half a squared times sine squared times omega T plus five. Let's see Plus cosine squared times omega T plus phi. Now we know from our trig identities that sine squared plus cosine squared is just equal to one. This is just going to be equal to one half times are spring constant times our maximum displacement squared while we have that. So let's just go go ahead and plug in our values here we have that are total mechanical injury. Energy is one half times are spring constant of 1 times our maximum displacement of 0.8 squared, which is equal to 0.4 jewels corresponding to our answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one

A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (e) the total mechanical energy of the glider at any point in its motion

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