a-spring-block-system-is-oscillating-without-friction-on-a-horizontal-surface-if-a-second-block-of-e

Question

A spring-block system is oscillating without friction on a horizontal surface. If a second block of equal mass were placed on top of the original block at a time when the spring is at maximum compression, which of the following quantities would NOT be affected?
(A) Frequency
(B) Maximum speed
(C) Amplitude
(D) All of the above quantities would be affected.

Nishant Kumar · Accepted Answer

So here we have a spring block system and a second block of equal mass is placed on the top of original block when it is at maximum compression. Now at maximum competition, it is at maximum distance from the equilibrium position, which means it is an at an amplitude, one of the extreme positions. So when the other block is put on this block, it will not change the amplitude because after the extreme position it moves to equilibrium position. And when the second block is put on the original block, its completion would not increase because of that, so the amplitude remains the same. Now Frequency of spring block system is given by ethical 21 by two pi mass, which is total mass of the system and spring constant. Sorry. Okay, it&#x27;s being constant by total mass. Now it&#x27;s been constant remains the same. But mass is changed when another block is added so frequency will be affected. Also, maximum speed is, uh V Max is equal to omega. Where is amplitude and omega is angular frequency and America is given by two by f. Now I&#x27;m produce. They&#x27;re not changed, but frequency changes, So maximum speed will also change. So after three options, frequency and maximum speed will change. But amplitude does not change. So the quantity remain unaffected. Would be amplitude. Yes. The correct option is C amplitude this complete solution. Thank you.

Keshav Singh · Answer

you can see the forces acting on Block B. In this diagram, the maximum acceleration off the oscillating system is given by a max and a Mex is equal to make a squid times eight Army guys angular frequency and is amplitude of oscillation. We know Omega is to buy F, so this can be written as full pie squid and the amplitude eight happens, the frequency squared. The friction force exists between the two blocs must be capable off accelerating block B at the straight. Thus, if Block B is about to slip, we can write an expression making the fiction on the fiction is at its maximum and this is equal to coefficient of static friction, U. S. And the normal force in Since the political forces are balanced, this is equal to us times in G. This is equal to mm times X elevation for pie squid A times of frequency squid as calculated above. So from here we can see we can find an expression for the amplitude eight. The amplitude using our extraction are expressions about you see what from us times G over full pi squared times have squid

Averell Hause · Answer

here for party. The amplitude is going to be the initial distance from equilibrium, so it was initially stretched eight centimeters. This means that the amplitude is going to be 8.0 times 10 to the negative second meters. This would be your answer for party four part to be then. Ah, we want to find the angular frequency for the angular frequency equaling the square root of the spring constant divided by the mass. And this would be the square root of the spring constant 425 Newtons per meter divided by 2.0 kilograms. And this is equaling 14.6 radiance per second. So this would be our answer for part B for part C. Then the frequency F is given as Thea think the angular velocity or the angular frequency divided by two pi. And so this would be 14.6 Radiance curse second, divided by two pi radiance giving us 2.32 for a second or 2.32 Hertz seconds to a native first for a part d, we want to find the period, so the period t is equaling. Two pi multiplied by the square root of the mass divided by the spring constant. This would be to pie multiplied by the square root of 2.0 kilograms and this would be divided by 425 Newtons per meter. And this is giving us 0.431 seconds. This would be our final answer for Part D. Um, we can also simply take the reciprocal of the frequency. Um, simply because the period is simply the reciprocal of the frequency. So either way, we can arrive 2.431 seconds to find our period. Four Part E. Uh, we know that here we&#x27;re going to find the maximum velocity for the block and so we can use the conservation of mechanical energy 1/2 times the spring. Constant times the amplitude squared equals 1/2 times m. The max squared. So essentially all of the initial potential energy associated with the spring will be transformed into kinetic energy. And so the maximum velocity is going to be equal in the square root of K squared, divided by the mass. And so this would be the square root of 425 Newtons per meter multiplied by 8.0 times 10 to the negative second meters quantity squared, divided by the mass of 2.0 kilograms. And we find that the maximum velocity is gonna be equaling 1.17 meters per second. This would be our final answer four Part E and four part F. We can say that to find the blocks maximum acceleration. We can first say that the maximum value of the acceleration would be equaling the negative amplitude times the angular frequency squared multiplied by co sign of two pi times the frequency times t time. And here then the maximum would be when co sign of this coastline to pie ft. When this value here is one, so the maximum would be simply a rather negative one. And so this would be amplitude omega squared. And so this would be equaling a times K over M, and we could find that the maximum acceleration. Then he&#x27;s gonna be equaling 8.0 times 10 to the negative second meters multiplied by 425 newtons per meter divided by 2.0 kilograms. And we find that the maximum acceleration. Is that equaling 17.0 meters per second squared? This would be our final answer for a part f That is the end of the solution. Thank you for watching.

Salamat Ali · Answer

Hey off the problem, the maximum can take. Energy is equal to maximum potential energy. From here we write equations for boat for spring we have can take energy. One word to every maximum square is equal to one over to K employees shoots square potential energy for spring from here. Then we&#x27;ll cancel out one or 21 or two on both sides and soulful. Do we? Then we were here. We get here is square root off. Okay, square. So it will come out here then Kay derided by him here. So? So I&#x27;m starting the value for amplitude, which is zero point 05 meter square, wrote off a K we have here is a 40 Newton per meter derided by mass which is a report one then this gives us they were lost You if one meter per second maximum velocity you bought me We were asked to find a frequency frequency busy pulled over on a spring Is one over two pi r squared off this spring Constant okay, divided by the mass. Then we can substitute the values for a K and them And uh for the two pi K we have is a 40 divided by again as a 0.1 mask. Cagey. That gives us the frequency off the 3.2 hearts. 3.2 hearts. So this was part Be in part. See, um the forces will be called forth. Restoring force will be equal to the weight off the right off this mass attached to a spring. And then we can replace restoring force by K X That is a cool toe mg from here x will be called to moss. Damn she divided by K then substituting the values we get. Tau your reproof press G 10 meters per second squared, divided by four d. This gives us the ex increase in lengthy zero point ah to five you major the natural lent off the spring would decrease by 0.25 Indy slipping the spring water curly the if the frequency remains unchanged

Matt Braby · Answer

So we&#x27;re gonna have a block here attached to a spring that was blocked. Pee with little block, Be on top of it, right, And it&#x27;s gonna be oscillating back and forth and there&#x27;s some friction here between those right on That friction is given by the 0.6. We know that in order for static friction to hold, then it must be less than us times the normal force and the force of friction is the only thing X accelerating Block B appear that no other forces acting on it that can cause the acceleration. And the normal force, of course, is balanced by its gravity. You will notice that those cancel and you end up with that. The acceleration of block B must always be less than in this case 0.6 times times the acceleration due to gravity. Well, the acceleration, though, due to the simple harmonic motion, is given by a Omega squared right. That&#x27;s the max acceleration. Right? And we&#x27;re separately told that frequency is 1.5 hurts, right? So Omega, with just two pi f is equal to three pie hurts, right? So the amplitude if I were to take this thing and make Plug it in right there. Right. We&#x27;re going to say that a Omega Square. It has to be less than or equal to me. Us times, Gene. Right. So a the amplitude has to be smaller than a certain amount. Too big, and you will get some slipping. Okay, so it&#x27;s 0.6 times 9.8, divided by three pi squared some nine pie squared and that amplitude ends up. You plug it in your calculator, you get about 0.66 meters.

Zachary Warner · Answer

for this question were told that we have a spring that&#x27;s attached to the floor face vertically upward. We then place a block on top of the spring. It&#x27;s not attached to the spring is just sitting on top of the spring. And, um, you were asked to find the frequency of the block, while the frequency of the block is related to angular frequency by which is omega by omega, divided by two pie. So that is the frequency of the block where omega is equal to the square root of Kate over him. So this is one over to pie times, the square root of Kay we&#x27;re in. So we just simply plug in our values for candy in here, and we find that the frequency of oscillation of the block and units of Hertz is equal to 2.66 hertz. For Part B. We were asked to find the maximum amplitude in which the block will disconnect from spring. So essentially, what it&#x27;s saying is what amplitude will cause the force of the spring pushing the block upwards to overcome the force of gravity. What The moment this happens, it means that the force of the spring has to be greater than the force of gravity. So the moment in which it&#x27;s happening is when the force of the spring is equal to the force of gravity. Well, the force of the spring was equal to K Times X, where X is the amount of compression. But in this case, the amount of compression is gonna be the amplitude right, cause it can&#x27;t go farther than the amount of compression. So the amplitude then would be the X value that we&#x27;re solving for. So this is equal to mess times gravity, where X is equal to the amplitude. So we&#x27;ll replace X with amplitude and solved for a theon platoon, and we find that it&#x27;s equal to mass times acceleration due to gravity, which is 9.8 meters per seconds squared, divided by K. You plug these values in and we find that this is equal to 0.35 meters, which we can box in Is your solution for part B

Problem

A certain theme park ride involves people standin…

Problem 5 Easy Difficulty

Answer

Related Courses

Cracking The AP Physics 1 Exam

Related Topics

Discussion

Top Educators

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

Robert Franek

Cracking The AP Physics 1 Exam

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Cracking The AP Physics 1 Exam

Robert FranekCracking The AP Physics 1 Exam

Robert Franek

Cracking The AP Physics 1 Exam