a-block-with-a-mass-of-4-mathrmkg-is-attached-to-a-spring-on-the-wall-that-oscillates-back-and-forth

Question

A block with a mass of 4 $\mathrm{kg}$ is attached to a spring on the wall that oscillates back and forth with a frequency of 4 $\mathrm{Hz}$ and an amplitude of 3 $\mathrm{m}$ . What would the frequency be if the block were replaced by one with one-fourth the mass and the amplitude of the block is increased to 9 $\mathrm{m}$ ?
(A) 4 $\mathrm{Hz}$
(B) 8 $\mathrm{Hz}$
(C) 12 $\mathrm{Hz}$
(D) 24 $\mathrm{Hz}$

Sachin Rao · Accepted Answer

Hi. In this problem, we have given a block with a mass of four kgs attached to a spring on the wall and oscillates back and forth with the frequencies for hedge and amplitude is three m. What were the frequency if the blog were replaced by you? Wanna fourth mhm. 1/4 the mass and the employee duties. That means, um prime is a call to m by four. And a prime is a cold Oh, mine meter. And it is a vlog increased to nine m night. So we have to find the new frequency f crime. So we know that frequency is equal to one of one to buy on the road gave I am so we can say it is proportional to and the road one day I am because K is constant so we can write f upon if prime upon f is equal to under road. Mm bye. Mm. Brian So this is a cold. Oh. Mm. Is M and M price M primaries and by four so f prime will be equal to two times of initial. So this is two and 24 That means it is equal to eight heart. So our options. Often we will be correct. Option. Mhm option. We use characters, and we know that frequency doesn&#x27;t depend on the aptitude. So it depends on the mask. So and we know that new masses one way. Fourth of initial. So this is a call to eight Hutch. The new frequency will be eight has

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.

Ze-Han Lee · Answer

right. So, uh, in this problem, we have a spring block system. So the mess of the block m equal. Full. Sorry. I mean, could turn kilograms. And we also know that that Ah, the amplitude of the oscillation, eh is Ah, two meters. And ah, Besides, we know that the maximum velocity for this oscillation is ah, full meters a second, so we can find out the total energy of the system. Right? So told energy of the system equal half m times V squid, where m is the mass of block and Avi&#x27;s the maximum velocity. So this equal half k time&#x27;s a squid. So this whole energy also equal to the maximum at maximum energy stored in the, uh in the spring, right? So Ah, we see that in this equation, we already know the mess. We&#x27;re in the velocity, and we also don&#x27;t Ah! Ah, the aptitude. So we&#x27;re gonna find out the K we&#x27;re capable 40 new temple meters. Okay. All right. So we can find out. Ah, the period of dissociation, which is to pie screwed em over. Kay. All right, so this gives you Ah, So am ease. Amis. 10 in a case 40. So this gives you ah, about pie. So the period is about ah three

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

Joanna Josey · Answer

for this problem. We&#x27;re are given the scenario, but there&#x27;s a 1.2 kilogram block sliding on a horizontal surface with a horizontal spring attached, and the spring has a spring constant of 480 noons per meter. Um, at time T equals zero in the block passes through X equals zero with a speed of 5.2 meters per second in the positive direction. Uh, and we&#x27;re being asked to find what is the frequency of the amplitude of the block and to write an expression for X as a function of time. So for the first part, finding the frequency, we&#x27;re gonna use the equation. Frequency equals 1/2 the square root of the spring, constant over the mass and so plugging in our values. We have 1/2 the square root of 480 noons per meter, divided by 1.2 kilograms, which gives us the frequency of 3.2 terms. So our frequency here ISS three 0.2 hurts now for the next part we&#x27;re trying to find what are amplitude is their exam. Um, and we can use the equation. X equals X m co sign of Omega T plus five on DDE. In this case, X equals zero and t also equal zero because we&#x27;re told that at times the a time zero the displacement was zero passed through position zero on DSO We can rearrange this equation in order to solve for if I so we want to solve for Fei here. So rearranging the equation we get that zero eggs will equals amplitude times co sign of five, um, and again solving for five we would divide by exam which would be zero, and the reverse co sign of zero equals fi which is which means that five equals pi over two. Now that we have fi, we can use the velocity equation where velocity equals omega times amplitude sign of omega T plus five um, velocity we have is 5.2 meters per second and for omega, we can solve for Omega, where omega equals the square root of K over M, which equals the square root of 4 80 divided by 1.2 because we had our mess and we were given the spring constant of 4 80 So that means omega equals 20 rads so we can plug that in. So we have 5.2 equals 20 times are X m sign of five guys again our time and serum and solving for amplitude Here we know that fi we can actually substitute out 54 pi divided by two. And if we solve for our x m, we get that X m equals zero point 26 meters. So that is our position. At near 0.26 meters. That&#x27;s gonna be our position or our amplitude for our oscillation. And then the last thing we&#x27;re asked to do is to write this in terms of time. So we have X equals 0.26 co sign of 20 times T minus pi over to, and that would be our expression.

Salamat Ali · Answer

Antique energy office. Spring mechanic Energy is equal to the potential energy office free so we can write a tactic. Energies one or two and we square from here. We will find amplitude so I can write potential energy off the spring is one or two k a square. So is the amplitude off this spring? So the amplitude came. You found by solving for a year canceling art one over two on both sides. Then the amplitude we get for a spring is in terms. Offer Mars and Velocity and Kay&#x27;s Every Square derided by K been constant. Maas we have here is a tree. We&#x27;re lost days of four. So four square derided by spring constant is a 10 squealed taking swirled off. This number gives us the answer is 2.2 meters, so our options sees the right answer.

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