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A 0.169-kg particle undergoes simple harmonic motion along the horizontal =-axis between the points z, 0.245 m and 381 The period of oscillation is 0.505 Find the f...

Question

A 0.169-kg particle undergoes simple harmonic motion along the horizontal =-axis between the points z, 0.245 m and 381 The period of oscillation is 0.505 Find the frequency; f, the equilibrium position; Im' the amplitude; A, the maximum speed Urnino the maximum magnitude of acceleration, Umix, the force constant; and the total mechanical energy, Etot .Etot

A 0.169-kg particle undergoes simple harmonic motion along the horizontal =-axis between the points z, 0.245 m and 381 The period of oscillation is 0.505 Find the frequency; f, the equilibrium position; Im' the amplitude; A, the maximum speed Urnino the maximum magnitude of acceleration, Umix, the force constant; and the total mechanical energy, Etot . Etot



Answers

Simple Harmonic Motion Find a function that models the simple harmonic motion having the given properties. Assume that the displacement is at its maximum at time $t$ = $0$. $$ \text { amplitude } 2.4 \mathrm{m}, \quad \text { frequency } 750 \mathrm{Hz} $$

Let's do part. A well angular frequency is equal to maximum loss. T divided by amplitude. We have maximum lost E, which is four point it and amplitude 0.25 4.8, divided by 0.25 his 19.2 ready. And for a second so angler frequency equals 19.2. Ready in for a second. Andi, which is approximately equal to 19 ready and or a second uh, part B Well, Time PDT is equal to, uh, Dubai, divided by F to buy do body by after this is the, uh time period or time toward its two pi, divided by Omega Time period is equal to two pi, divided by omega and hence to buy divided by 19.2, which is equal to 0.33 seconds. And Farsi Well, here we have acceleration, right, so X relation is equal to eight times Omega Square, and this can be region is a times Omega Times omega and eight times omega is maximum velocity, right times omega. Now this is equal to 4.8 4.8. Right is the maximum lost into mega, but we'll make a C 19.2, and therefore we have X relation is equal to, UH, 92 92 meter per second square, 92 meter or second square.

Well, let's do party. We know they'd. Maximum acceleration is equal to eight times will make us. We're We raised the amplitude in Omega's the angular frequency, and a maximum velocity is equal to eight times omega. No, let's divide them. So maximum acceleration, divided by maximum lost e, is equal to a omega Tom Hey Omega Square into a times omega divided by eight times omega. And therefore the ratio between maximum acceleration and maximum velocity is equal to omega, right? No, we can find out omega well. Omega is equal to 3.1, divided by at 1.4, and omega is about to a point to radiant or a second still going to radiant or second, no, we find out time, period. So Time PDT is equal to do by divided by omega and to buy, divided by omega to make a two point 214 which is equal to two point eight seconds. And Marcy, let's find a damn platitude. Will Amplitude is given by maximum velocity divided by, um, angler frequency, but maximum. We've lost these one point for an angle. Frequency is 2.214 right, and therefore amplitude equals 0.63 meter

And this problem were given the amplitude, frequency and displacement at Time T equals zero and were asked to find a function to model that. So the first thing we can do is pointing a Our amplitude there are implicated is gonna be 6.25 next week and use a relation between frequency and Omega to find Omega. So every plug in 60 for frequency in Seoul For omega we get that omega equals 120 pie. Next, running user information about displacement to find fee. So first, we're going to plug in our maximum height of 6.25 for why we're in a plug in our impl itude 6.25 Our sign Omega T is going to go away because T is equal to zero. They were gonna have negative fee. You can divide the 6.25 over, which is one sign negative fee. Whenever you saw for fee, you gets it that is equal the negative pie halves. Next, we just have to combine everything together. Their final answer of wife was 6.25 sign 120 pi plus by halves. And this is a plus here because are negative pie halves turns the original negative

So we're gonna be looking at problem 21 of Chapter 17 of the Physics 5th and we have a particle undergoing some simple harmonic motion. Um and it gives us an amplitude and a maximum acceleration. And the party we need to find the period while we're just going to use a max is equal to omega squared A where a is the amplitude. So omega squared is 7.93 times 10 to 3. To put it in meters per second squared to this is please meters that can square and divide that by 1.8, 6 times 10 to the -3, which is our amplitude in m. Um So we get omega is equal to the square root this and we get omega is equal to 2000 and 60 radios per second to three sig figs And T is equal to two pi over Omega. So we get obviously to pi over 2060 and we get 3.04 times 10 to the -3 seconds. Um or each for the time period for part B we need the maximum velocity. Well this is just very simple. We're going to use the equation V max is equal to omega times of the amplitude, So 2000 and 16 times by 1.86 times 10 to the monastery. And we get 3.84 m per 2nd, 2366. And from part C we need to calculate the total mechanical energy. Well the total mechanical energy will say M E is equal to k e s p E. But we know at at the point when um the velocity is at its max, at v max, K E is equal to half mv squared max and p E Is equal to zero. There's no potential energy. So our total energy m e is equal to a half m b squared max, half times 12.3 times 3.4 squared. And that gives us a final answer of 19.7 Jules to three significant figures.


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