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8. A mass at the end of a spring moving in simple harmonic motion has an amplitude of 0.0450 m and maximum acccleration of 40.0 mls2_ What is the frequency of the s...

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8. A mass at the end of a spring moving in simple harmonic motion has an amplitude of 0.0450 m and maximum acccleration of 40.0 mls2_ What is the frequency of the system? 187 HzE] b. 1.80 Hz 14.9 Hz 4.75 HzEl 11.9 Hz9.A simple pendulum has period of 5.00 What is the pendulum length? (g 80 m/s2) b. 0.161 m 0.0403 m d.19.5 m 6.21 m 24.8 at 130 Hz: and 390 Hz, one of these resonances being the 10. The air in a pipe resonates resonances are between the (wo given fundamental. If the pipe is open at b

8. A mass at the end of a spring moving in simple harmonic motion has an amplitude of 0.0450 m and maximum acccleration of 40.0 mls2_ What is the frequency of the system? 187 HzE] b. 1.80 Hz 14.9 Hz 4.75 HzEl 11.9 Hz 9.A simple pendulum has period of 5.00 What is the pendulum length? (g 80 m/s2) b. 0.161 m 0.0403 m d.19.5 m 6.21 m 24.8 at 130 Hz: and 390 Hz, one of these resonances being the 10. The air in a pipe resonates resonances are between the (wo given fundamental. If the pipe is open at bothan.owonnyaeyc the two given ones? is closed at one end, how many resonances are between ones, and if the pipe 0; closed: open: closed: b. open: 2; closed: 0 open: open: 1; closed: 0 EDa



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21. Massless Spring A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position $y_{1}$ such that the spring is at its rest length. The object is then released from $y_{1}$ and oscillates up and down, with its lowest position being $10 \mathrm{~cm}$ below $y_{1}$. (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is $8.0 \mathrm{~cm}$ below the initial position? (c) An object of mass $300 \mathrm{~g}$ is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) Relative to $y_{1}$ where is the new equilibrium (rest) position with both objects attached to the spring?

Hi here it is. Given a block of mars m kg suspended from the spring office spring constant Cape. It's having the expansion of .025 m. Sure we can find its natural frequency at equilibrium MGB K delta X. So the value of you will get MG upon their tax. Natural frequency is given by a root of Do you buy em? That is MG upon delta X upon. I am so omega not will be g upon date. Tax GS 9.81 extension is 0.25 m. So natural frequency of the loaded spring 19.81 Already in for 2nd. Second part. Because the top of the spring is oscillating since top of the spring is oscillating. So the spring force on the mask, mine escape X minus dick. So equation of motion of the mosque given by Thank you. Yeah. An external dot. It's called to mine escape x minus I minus b. Expert. So this ultimately could be at an acceptable dot gamma x dot Because omega not square eggs, it's cool to omega notice squares. I Mhm Okay. Yeah, course up. Okay, we got to mhm If mega not try not physical to have, not by him, I'm pretty good. Can be written as omega noticed cause I not upon a group of omega not square minus omega square. Let's go, come on you guys square. This could be reduced to we don't notice where they not upon comma. Omega not that is I not omega not upon comma, that is I not cube. Why did you does your upon into 15 so you will get .015 m or 15 cm. Mhm. See bug the main power input. It can be not an sp me mega will be no comma. Have not squared upon Who have four times. Oh, Megan not minus omega square, let's go my square please. So be average. You can right a mega not into America not to the powerful. They not expect every square upon two Qm into going up on four times. America not minus omega is square plus mega not square by cube. Mhm. And finally it can be a little less half. They not square. We're not to the power amp We've done upon four omega not minus omega squared less mega not the power omega not square by cube. No using Omega is called Omega not one plus Egypt he had he to having the value 102 So bridge power you will get mhm holes they noticed square and we can not kill him Q upon for eight I square upon Q. B squared plus one. Now substituting the value. You will get the answer. Yeah yeah yeah into one upon For it to square on 0 to square upon point doodle doodle 15 square. But yes, but On simplifying it, the power will be .086 ports. That sort. Thanks for watching it.

Hello students. In this question we have an object of mass M equals to 0.2 kg is hung from a spring. Who's the spring constant? K. It is equals to 80 newtons per meter. And the object is subjected to a registry force given by F equals two minus B. P. We are V. Is the velocity in meters per second. So for the part A we have to set up the differential equation for the motion of free oscillation, free oscillation of the system. So hence we can write that evacuation of motion is acceptable. Dot plus gamma extort plus omega not is square molecular B X. It is equal to zero where gamma it is equal to be by M. And omega notice square, it is equal to K by M. So this is the differential equation for the required free oscillation. Okay, now, for the part B we have to calculate the damping constant be if damned frequency omega is equal to undergo three by two of the condemned frequency. that is omega not. Okay, so from here we can write that the omega square, it is equal to three x 4 of Omega Notice Square. So hence we can write that uh this omega squared, this is equal to omega notice square minus gamma square by four. So omega square which is equal to this three by four times of omega notice square. So from here after solving comma it is equal to omega not Okay. And be it is given by Emma claire by omega not or it can be written as underwrote of cable. Am under KMc. Okay. Which is given as 18 newton per meter and msg 0.2 and this under. So from here we get be it is equal to four newton's second per meter. So this becomes the answer for this part be okay now for the part C in which we have to calculate the value of the system and by what factor the amplitude of the oscillation is reduced after 10 complete cycle. Okay, 10 complete cycle. Okay, so first of all calculating the value, so Q value, it is equal to omega. Not ever be gamma. So omega not which is equal to the root of K by M gamut it is equal to Bebe. I am so one by gamma will be equals to em by B. Okay, so we can write that under root of Okay, which is equals two. Okay, which is given as Spring constant, which is given as 80 and masses given as 0.2, multiplied by 0.2. There were baby which is obtained as four. So from here we get you, it is equal to one. Okay, so this is the value of the system. Okay, now, solving the amplitude a. So evacuation of amplitude a. And develop a. It is equal to E. To the power minus gamma T. N. divided by two. Or it can be written as into the power minus B. 25 10 developing too By two M, manipulated by Omega. Okay, so substituting the values so we will get that This value will be equals 2 to the power minus B, which is obtained as four more popular by two pi and They were big too and mm 0.2 and Omega which is equals two omega which is obtained as a Omega which is obtained as three x 4. three Under 3 x two times of Omega. Not okay. And we can substitute the values so we will get this value as uh omega. Not this can be substituted. So this be can be also replaced. Okay, so be can be written as in the term. So we can write it this equation in the other form. So it was the power minus. And this be can be written as omega not M. Pie. And developed by Mm Omega not manipulated by under Rule three x 2. So from here we will get that this value is equals two to the power minus cooper and they will be under a tree. So and is given as 10 cycles so 8 10 divided by a. This is equal to the to the power minus two pi molecular by 10, develop a under three. So from here after solving we will get that 8 10 divided by a. It is equal to 1.76 molecular by 10 to the power minus 60. So this becomes the answer for the part C. Okay thank you.

So for the first part, we're gonna use a question 17.39 to find the frequency. So according to this a question, we have frequency caldo and times the speed off sound over four times the length off the tube and since via the yard, finding the fundamental Modoff vibration. So we use an equal to one and speed of sound is 2 43 meters a second. Then we have four times the length off that too. And this gives the frequency to be 71.46 hoods now coming to part B. So for the wire, um, it will use a question 17 point 53 again. Now for the lion we have after she will do and the via over guys off a lion. And this is equal. D'oh one always do a liar. Jj, times towel overs new route over where mu is moss off the lion over length off the wire. Some use the union moss density off the lion on recognizing that F is if equal to f dash. And it's right that because both the wire on air in the tube vibrated but the wire on the air in the cube vibrate at the same frequency, so this condition should be satisfied. Now, we love these two expressions, so we have already calculated s so we don't need a tow used this expression Now, over there on, we will use this expression for if Nash So we have one over Dries off l a liar Times go over me and this will give us stall to be equal to on I forgot to say How is the tension in the wire? So therefore we plucked and Time Street equal dorms the ivy, equal stool attention or word the Linnean moss density on for attention We used how on new Like I have already said Newest Alina mass density on DDE Also, since this is also the fundamental frequency that we're solving for so and it's taken to the one. So now the so $4000 is twice off and love wire times If who's got times? Most of the lion over love the virus On simplifying this expression we get out, Toby flown f squared moss off the wire dies, went off the wire and plugging the values. So we have moss iss. I won six thanks into the by minus three in cages, then lent off the wire, noticed that here we had the length off the tube, substituted and hearing it the lint off the wires. Don't get confused on DDE. Mix up the values. This game started to be 64 point Dayton says he have now the tension in the wires.

So we know it's ADA. The angle measure will be equal to the aptitude times the exponential function of negative gamma T times co sign of omega prime tea. Uh, the exponential function simply means that whatever is inside the parentheses is is the power with the base of E. Ah, we know that a the amplitude is 15 degrees and that tea is eight seconds, so eight seconds have elapsed for party. We know that 5.5 degrees will equal 15 degrees times the exponential function of negative eight 0.0 seconds times gamma. And so solving for gamma gamma will simply be equal to negative one divided by eight times the natural log of 5.5, divided by 15. And so gamma is going to be equal to positive 0.1254 per second for a part B. They want us to find the period so we can say that omega not the original, uh, angular velocity would be equal to MGH over the moment of inertia. And this is gonna be equal to, uh, MG times one over to l divided by the moment of inertia. In this case, it would be 1/3 ml squared and this is giving us the square root of three G over to L. Now we can find the formula for the damp frequency, the damned frequency where we can say omega prime would be equal to the square root of the original angular velocity squared minus gamma squared. Also, this 1/2 power. So this is going to give us the square root of three G over two times l minus gamma squared and be solved. So, uh, Omega Prime would be equal to three times 9.8, divided by two times 20.85 meters, and then this will be minus 0.1 to 54 squared. And this is giving us 4.157 radiance for a second. So keep track of that and the T prime. The period will simply be equal to two pi divided by omega prime. So this would be to pie a rather divided by 4.157 and we find the period to be 1.51 seconds. So that's our final answer. This being our final answer for part A. And so for part C part see, simply wants us to find the ah, the time it takes to reach seven her half of the amplitude so we can say 7.5 degrees equals 15 degrees times e x p the exponential function of negative gamma times. T therefore, t the half. The ah t sub 1/2 would be equal to Ellen of 15 divided by 7.5 or Ellen of two, divided by 1/2 15 7.5 2 20.1 to 54 And this is giving us 5.53 seconds to reach half of the amplitude. So this would be our final answer for part C. That is the end of the solution. Thank you for watching.


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