When analyzing motion physicists like to look for quantities...

When analyzing motion, physicists like to look for quantities that remain constant. The amplitude is one of those. Let's measure another one: 9. Click and drag the stopwatch out for use. 10. With the speed set to "slow", measure the amount of time needed for the spring to complete one full oscillation. This is called the period of the motion of the spring. Record its value below: Period= 84 s For various reasons, physicists often rewrite the period into other quantities. Calculate and record each in turn below: 11. Determine the frequency of the spring's motion via the following calculation: = Where fis the frequency and T is the period. Note the units of frequency are oscillations per second, or hertz (Hz): |- 84 Frequency = 1.19 12. Determine the angular frequency of the spring's motion via the following calculation: ω= 2π == T Hz : 2πf Where is the angular frequency. Either calculation (with T or with f) will do. Note the units of angular frequency are radians per second: 20 rad/s Angular frequency = 7.

Transcript

00:01 In this problem we are given the displacement function for an object that is suspending vertically at the end of a spring as follows.

00:11 Y of t equal to some amplitude times cosine of omega times t.

00:19 Here omega is a constant that depends on the mass of the object and the stiffness of the spring and it also means the angular frequency of the speed.

00:32 Periodic motion of this harmonic motion.

00:35 Okay with that with this function we are going to do some stuff over i think four parts so let's get started with the first one.

00:46 In this first part we are going to show that this function this solution of motion satisfies the following differential equation the second derivative of y equal to minus omega square times y so to do that we are going to take the second derivative of this function and try to see if this is equal to some constant times the function itself so let's get started with the derivatives the first derivative is minus omega a sine omega t and the second derivative is minus omega squared a cosine omega t but this is just the function itself so we have minus omega squared time why and this i show that the given function satisfied this differential equation indeed okay next part here we will show that the period named at the time required to make one complete oscillation can be written as t equal to two pi over omega to do that we are going to on the try to understand the very definition of this omega factor.

02:12 Omega is one revolution or one complete oscillation in t amount of time, let's say t seconds.

02:28 But one revolution is just two pi radiance, that is the circumference of a circle, of a un circle, two pi radiance over t seconds, and with this relation only, we see that t is equal to 2 pi over omega.

02:51 Well, we are measuring t in seconds, most probably, and omega in radiance.

02:58 Okay, next part.

03:02 Now there is this frequency, f of the vibration, which is just the number of oscillations per unit time.

03:10 With that, we are going to find a relation between f, this frequency, this frequency, and the period t.

03:17 Okay, again, let us try to understand the definitions of these variables or parameters.

03:24 Period is just the total time that it takes for an oscillations...

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