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A body-spring oscillator oscillates horizontally on africtionless surface. The mass of the oscillator is500 g and the spring constant is 200 N / m. The amplitude of...

Question

A body-spring oscillator oscillates horizontally on africtionless surface. The mass of the oscillator is500 g and the spring constant is 200 N / m. The amplitude of theoscillation is 1.0 cm. Calculate: (a) the number ofoscillations made by the oscillator in 1.0 min; (b) the speed ofthe oscillator as it passes through the center of theoscillation.

A body-spring oscillator oscillates horizontally on a frictionless surface. The mass of the oscillator is 500 g and the spring constant is 200 N / m. The amplitude of the oscillation is 1.0 cm. Calculate: (a) the number of oscillations made by the oscillator in 1.0 min; (b) the speed of the oscillator as it passes through the center of the oscillation.



Answers

A simple harmonic oscillator consists of a block of mass 2.00 $\mathrm{kg}$ attached to a spring of spring constant 100 $\mathrm{N} / \mathrm{m}$ . When $t=1.00 \mathrm{s}$ , the position and velocity of the block are $x=0.129$ $\mathrm{m}$ and $v=3.415 \mathrm{m} / \mathrm{s}$ (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at $t=0 \mathrm{s} ?$

For this problem on the topic of energy conservation, we are told that a one kg mass is suspended vertically from a spring. The spring has a spring concert of the 100 newtons per meter. And is the oscillations have an amplitude of 0.2 m at the top of the oscillation. The masses then hit so that it moves instantaneously downward with a speed of one m per second. We want to find the total mechanical energy of the system, the speed of the mass as it is moving as it crosses the equilibrium point and its new amplitude. Now the total mechanical energy before the hit is E. And E is equal to half K. He squared for the oscillations. After they hit. The new mechanical energy, we'll call it e. New is equal to half Okay, A squid plus a half M. V push squared where we pushes the speed with which the masses pushed. Now the new speed equilibrium can be found by setting A. Is equal to zero and we have half. Um the new squared is equal to he knew. And the new amplitude two can be found by setting V is equal to zero and so we get a half. Okay, A new squared is equal to the total mechanical energy. And so for part A the total mechanical energy he knew is equal to half K A squared plus a half M. V push squid and this is a half times 100 newtons per meter Times the amplitude of oscillation, 0.2 m. And that's squared plus a half times the mass of one kg times. It's The speed at which it was pushed one m per second squared. So the total new mechanical energy is 2.5 jewels for part B. We want to find the new speed. So from above equilibrium, the new speed, the venue is the square root of two times the new mechanical energy divided by mm. So this is the speed as it passes the equilibrium point, which is the maximum speed. This is the square root of two times 2.5 jewels, divided by the mess one kg. This gives us the speed to the equilibrium point to be too 0.24 meters a second. And lastly the new amplitude. A new can be found by the square root of two times the new mechanical energy divided by the spring constant of the spring. Kay again these values are known. This is the square root of two Times 2.5 jewels, divided by 100 newtons per meter, which gives the new amplitude of oscillations to be zero point 2 to 4 meters, Which is 22.4 centimetres.

Hi, everyone. This is the problem based on concept of frequency, of oscillation, of Lord listening. If I spring office brings question K loading by the body and then its frequency is one upon goodbye route of Cuba. I am? No. In the first quarter, it is given a spring question of the spring is 2.5 Newtons per meter mast loaded? Having the value. I'm one kg your frequency, You will get one upon to five. Okay, 2.5151 So it is 3.796 hearse. Be part. All right. Sorry. Amplitude of oscillation. You will calculate. Mhm xD square. Blessed beauty square upon group F for displacement minus five centimeter velocity is given from the centimeter per second and frequency We have measured 12796 The on solving it application. You will get 6.4 centimeters. Yeah, Pardon me. Mhm applying contribution of energy. Okay. Uh mhm for initial position. Five centimeter and final position. Three centimeters. Behalf and the I square half K X. I split half and we finally squared plus half. Okay, accept no substitutes. The volume to calculate your final moss is quite What be well is 20 centimeters. That is fine too. He is 2.5 and 2.5 He put off. Okay? Yeah, I must point what? We have to spend half 0.5, right? And two. Why? Angina? Three. It's square. Okay on solving it. Final velocity. You will get from P 8. 20 centimeters per second there, so thanks. Part of

Solving party of this problem here. The speed can be written had B. X. Is equal to Underwood. Okay but I am multiplication is square minus X squared describing both sides. So I can write the expression like this on. For the simplification I can write the value of A. Is equal to Underwood and by Kay multiplication we exist squared plus X squared. So I will just put the value in the next step. We're just putting the value in this expression. The expression can be written it understood .60 Kg by 14. Newton permitted multiplication .95 m/s. Holy Square Plus .22 m square which is equal to 0.295 m as the amplitude. Now solving part B. I can ride. The value of is equal to one x 2K. A square which is equal to one by two. Multiplication, 14 U. Turn parameter multiplication. 0.295 m police square. On solving it further. Finally I can ride evaluate just look at it carefully. 0.609 jewel as total mechanical energy. Now solving part C. The year I can write the value of P. X. Is called to under route. Hereby AM multiplication is square minus X. Is square. To just putting the value here I can I. D express a net under the hood. 14 x 0.60 multiplication 0.295 meter holy square minus 0.11 m holy square Which is equal to 1.32 m pulse again as our answer.

So here for part A, the period T is simply the reciprocal of the frequency. This would be equaling 2 1/2 20.0 hertz. This is giving us then point 50 seconds. We have then part B, the angular frequency equaling two pi times the frequency This would be going to two pi multiplied by 2.0 hertz, giving us four pi radiance per second. Now we're going to use energy conservation 1/2 times K A squared plus 1/2 times K X initial squared. Rather, this is Equalling plus one, huh? M V initial X squared. So essentially the total potential energy that can be stored is equal to the currently stored the current stored potential energy in the spring, plus the initial kinetic energy. And so ah, solving. Then we can eliminate that 1/2 and we find that then k a squared equal in K ex initial squared plus m e initial x squared Okay, is equaling. Then, um, Omega squared. We can solve for a and is equaling done 5.54 centimeters. Again, this is we're using ex initial equaling negative 5.0 centimeters and we have the initial explosively equaling negative, 30 centimeters per second. And so we can say that then we also know the mass equaling point 200 kilograms. So four part D, then we can calculate the face constant, a co sign of find, not Equalling X initial equaling than 5.0 centimeters. And we can say that then find not the phase. The face constant is equaling Arc co sign with 5.0 centimeters divided by 5.54 centimeters and this is giving us 0.45 radiance. So this would be the face constant, the maximum speed for part E. This would be going to the amplitude most played by the angular frequency. This would be equaling two 5.54 centimeters multiplied by four pi radiance her second. And this is giving us 8.8 right there. My apologies. 70 centimeters per second rounded to one significant figure. The maximum acceleration. Then this is equaling omega square times A. This is equaling. Then for pie, radiance for second multiplied by the maximum velocity or 70 centimetres per second. And we find that this is Equalling 8.8 meters for a second squared So for part G, then the total energy would be equaling 2 1/2 times the mass times the maximum velocity squared. And this is giving us 1/2 multiplied by point 200 kilograms multiplied by points 70 meters per second. Quantity squared. And this is giving us 0.49 jewels. My fault. It's 0.0 for a night. It's only 10 my policies and then we can say for part h, we can say that the position at T equaling 0.4 seconds, this would be equal to 5.54 centimeters multiplied by co sign of four pi radiance per second multiplied by point for seconds, This would be plus 0.45 radiance. And we find that then the exposition at T Equalling 0.4 seconds. This is gonna be equal to positive 3.8 centimeters. That is the end of the solution. Thank you for watching


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