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(III) A glider on an air track is connected by springs to either end of the track (Fig. $39 ) .$ Both springs have the same spring constant, $k,$ and the glider has mass $M .$ (a) Determine the frequency of the oscillation, assuming no damping, if $k=125 \mathrm{N} / \mathrm{m}$ and $M=215 \mathrm{g} .$ (b) It is observed that after 55 oscillations, the amplitude of the oscillation has dropped to one-half of its initial value. Estimate the value of $\gamma,$ using Eq. $16 .(c)$ How long does it take the amplitude to decrease to one-quarter of its initial value?$x=A e^{-\gamma t} \cos \omega^{\prime} t$

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a. $$5.42 \mathrm{Hz}$$b. $$0.068 \mathrm{s}^{-1}$$c. $110 .$ oscillations

Physics 101 Mechanics

Chapter 14

Oscillators

Motion Along a Straight Line

Motion in 2d or 3d

Periodic Motion

Cornell University

University of Michigan - Ann Arbor

McMaster University

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

02:18

In physics, an oscillation is the repetitive variation, typically in time, of some measure about a central value or between two or more different states. The oscillation may be periodic or aperiodic.

09:06

#1) An air-track glider is…

01:40

III An air-track glider is…

03:21

A $175-$ glider on a horiz…

01:52

A 175-g glider on a horizo…

03:51

A 175 -g glider ona horizo…

06:56

A simple harmonic oscillat…

02:31

A $250 \mathrm{g}$ air-tra…

06:53

A 10.6 -kg object oscillat…

A mass on the end of a spr…

04:28

$\bullet$ A 0.500 kg glide…

so for party. We're trying to find the frequency so the period is gonna be to pie times the square root of the mass divide by case of one plus case of two. Now, uh, this is going to be equal to two pi times Theme ass divided by two K. Because again, case of one equals case up to s so we can say that the frequency is going to be able to one over to pie times the square root of two k divided by M. Uh, we can say that this is gonna be equal to the square root of K over to pi squared M, and we can say that the frequency is I'm going to be 125 Newtons per second and then this will be divided by two pi squared times the mass of point two 15 kilograms, and we find that the frequency is approximately 5.43 hurts. So this would be your answer for party and then for part B, we want the gamma won't affront gamma so X would be equal to the amplitude times the exponential function of negative gamma times T Times Co sign of omega Prime times T. And so the maximum ecstasy, the maximum displacement would be equal to the amplitude times the exponential function of negative Gamaty. Ah, Now here we know that this is equaling a over to, and this is happening on Lee after 55 periods. So we can say that 1/2 of the amplitude with equal to a exponential of negative gamma t and then, uh, we can find that gamma is gonna be equal to Elena to over 55 times t the period in this case. So this is gonna equal the frequency times, Alan of two divided by 55. So gamma is gonna be equal to the frequency of 5.43 hurts times Ellen of two. And then this would be divided by 55. And we find that gamma is gonna be point 0684 for a second. So this would be your answer for part B and then port. See, uh, we are simply trying to find t. Yeah, so X max would be equal to a exponential of negative gamma t. Uh, and we're trying to see how long it takes to reach 1/4 of the amplitude. So T is gonna equal Allen of four divided by gamma. And this would equal Allen of four, uh, divided by 0.684 her second. And we find that tea is gonna equal 20.3 seconds. And again, it takes 24 20.3 seconds. In this case, it'll complete 110 oscillations. So this would be our final answer for apart. See, that is the end of the solution. Thank you for watching.

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