A block on a frictionless table is connected as shown in FIGURE P15.75 to two springs having spring constants k₁ and k₂. Find an expression for the block’s oscillation frequency f in terms of the frequencies f₁ and f₂ at which it would oscillate if attached to spring 1 or spring 2 alone.

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A spring mass oscillator is used in a mechanical clock to be exhibited during the school science fair. It consists of two springs with a spring constant K one and K two where K one equals K two divided by four. And the Bob of mass M as shown below, derive an expression for the Bob's oscillation frequency F in terms of the frequencies F one and F two at which it would oscillate fa F attached to spring one or spring two alone. So that is our end goal. OK. So we're given some multiple choice answers here. Let's read them off to see what our final answer might be. A is F is equal to the square root of 1/5 multiplied by F one B is F is equal to the square root of one divided by five, multiplied by F one C is F is equal to the square root of 1/ multiplied by F two. And D is F is equal to the square root of one divided by five multiplied by F two. OK. So first off, let us make the following assumptions. The mass of the spring is Neil, which it can be ignored. Compared to the mass of the bomb, the air resistance can be ignored. The amplitude is relatively small such that it does not deform the springs. So to begin, let us create a diagram to help us visualize the forces acting on the spring mass system. And the change in the length of each, let us also draw free body diagram of the Bob itself. So here we go. So we take the diagram that was given to us in the problem itself originally. And then over here to the left, we added a lot of more detail to it. It shows that the force of one on C is an arrow is the red arrow pointing down. And that FC on one is pointing up and then F two on one is pointing down and then F one on two is pointing up and FM on two is pointing down and then F two on M is pointing up. And then for our free body diagram, we have it pointing up for the F 21 M like we mentioned before. OK. So at this stage, we need to find the net force acting on the Bob using the second law of motion. Let us determine or let us write the net force on the bob in terms of magnitude. So let's do that. So the force two on M is equal to negative A two multiplied by delta Y two. Now let us consider the third law of motion which states that vector F 21 M is equal to vector M on two. OK. So if we consider the net force acting on spring two, we can write that vector FM on two minus vector F one on two is equal to zero. Thus, we can take it even further to write vector M on two is equal to vector F one on two. OK. So now we can combine the net force equations. We found to write at vector F 21 M is equal to vector FM on two is equal to vector F one on two which we can go on to simplify and write that vector two on M is equal to negative K two multiplied by delta Y two which is equal to negative K one multiplied by delta Y one. Or we could write negative K two multiplied by delta Y two is equal to negative K one multiplied by K. So OK. So negative one, sorry, negative K one multiplied by Y delta Y one. OK. So now we need to determine the net vertical displacement of the two springs. So recall the equation for the net vertical displacement is delta Y is equal to delta Y one plus delta. Y two, which is, we need to go on to write that delta Y is equal to negative F 21 M divided by K one plus my negative F 21 M divided by K two. So we need to simplify. So delta Y is equal to negative F two on M multiplied by one divided by K one plus one divided by K two, which we need to simplify it one more time. So delta Y is equal to negative F 21, M multiplied by K one plus K two divided by K one multiplied by K two. Fantastic. So now we need to express the net force on the ball based on the net displacement of the two springs and on the spring constants. So vector F two on M is equal to delta Y or negative delta Y divided by a one plus K two divided by K one multiplied by K two, which is equal to negative delta Y multiplied by K one multiplied by K two divided by K one plus K two. So we can then write that vector F two on M is equal to negative K effective multiplied by, or I should say efficient or KF multiplied by delta Y where KF or K sub sort of F efficiency is equal to K one multiplied by K two divided by K one plus K two. So we can write that F two on M is equal to KF KF multiplied by delta Y OK. So we're going to switch gears a little bit here. We're still, we're chugging right along. OK. So now we need to express the angular frequency of the Bob in terms of the angular frequencies of either the spring of either spring acting on the Bob alone. So we need to focus on the angular frequencies of either spring. So spring one or two acting on the bolo. OK. So Omega represents the angular frequency is equal to the square root. OK. Effective divided by M, the mass is equal to the square root of one divided by M multiplied by a one multiplied by K two divided by K one plus K two. But we need to simplify this. So Omega the angular frequency will be equal to the square root of K one divided by M multiplied by K two divided by M all divided by K one divided by M plus K two divided by M. But we must recall a note that let's make a little side note here that K one is equal to K two divided by four. So we can write that Omega is equal to the square root of K two squared divided by or M squared all divided by M multiplied by K squared plus or MK squared divided by four M squared, which is equal to when we do some simplifying. So K squared or K two squared divided by five multiplied by M multiplied by K squared is equal to the square root of K two divided by five multiplied by M. So Omega is equal to the square root of K squared. So we do even more simplify to get to this point divided by five, multiplied by M is equal to the square root of Omega two squared divided by five which is equal to the square root of one fifth omega squared. So finally, we could solve for our final answer, we can now find the frequencies F. So note that the angular velocity or angular frequency is equal to two pi multiplied by the frequency. So when we rearrange this to sol for the frequency we get F is equal to omega divided by two pi. So the frequency is equal to one fifth multiplied by omega squared divided by two pi. So the frequency is equal to the square root of one divided by five, multiplied by two pi multiplied by F two, divided by two pie. So when we simplify, we get that F is equal to the square root of 1/ multiplied by F two, which is our final answer. OK. So let's go look at the multiple choice answers at the beginning to see what the correct answer is. So that means the correct answer has to be the letter CF equals the square root of 1/ multiplied by F two. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.