A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height 3.0 cm above the top of the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 10 cm. What is the oscillation frequency?

Video transcript

Everyone. Let's take a look at this practice problem dealing with a simple harmonic motion. So in this problem, we have a vertical spring with the bottom end fixed to a rigid surface and the upper end is covered with adhesive. The spring has a natural length of al knot and a small sphere has dropped from rest um five centimeters above the upper end of the spring and adheres to the spring system formed by the spring of the sphere oscillates with a maximum displacement of eight centimeters. And the question was just to calculate the angular frequency of the sphere. We're given four possible choices. Choice A is seven radiance per second. Choice B is 12 radiance per second. Choice C is 15 radiance per second and choice D is 21 radiance per second. So let's start off by drawing a figure of our problem. So I'll draw a circle for the ball that's gonna be above the spring, which I'll draw below it. And that spring is attached to a surface which will draw is a horizontal line. Now, we need to identify a couple of quantities in the setup. So we're gonna call the distance from the ball the spring as H we call the distance from the uns stretched length of spring to the new equilibrium point for the oscillations to occur or call that distance X. And weren't called the distance from the new equilibrium point to the maximum displacement as a which would be our amplitude. So we're asked to calculate the angular frequency of the sphere. And so let's recall our formula for the angular frequency for a mass in a spring. There are angular frequency omega can be equal to the square root of K divided by M K is the spring constant and M is the mass. Now, I don't know either of those two quantities. The only thing I was given were distances. So let's look at another um area where we could possibly relate the mass and the spring constant to a distance. And that's the point where we're at equilibrium where the weight is equal to the spring force. So in that case, our weight MG is going to be equal to the fourth due to the spring, which is gonna be KX at this point um can actually solve for the ratio of K divided by M, we'll have K divided by M and that's gonna be equal to G divided by X. So what that means is that our angular frequency could be written as the square root of G divided by X. So X here is the distance that the spring is compressed the new equilibrium point NG is acceleration due to gravity. So really now I need to do is find how the distance X relates to the H and the A and for that, we're going to look at the energy conservation. So initially, I have all gravitational potential energy. And so let's first define a zero point for gravitational potential energy. And we'll set the Y equal to zero point to be when the spring is fully compressed to its maximum distance. So doing that, we can set the gravitational potential energy, which if you recall is just mGy. And so that's gonna be MG and for the Y value that's gonna be H plus X plus A. So that's gonna be the start uh the ball starting why value? Now when it's fully compressed, I have no more gravitational potential energy. I have no kinetic energy since it's momentarily stopped. But I do have spring potential energy. So that's gonna be equal to. This brings potential energy is one half K modified by the compression length, which in this case is X plus A and that quantity gets squared. So if I look at this, I now have a relationship between XH and A but I still have this am in this um formula. So what I'm going to do now is divide both sides by the one half K. And in doing so I'll get two, then I'll have MG divided by K and then I'll have that multiply the H plus X plus a quantity. And on the right hand side, I'll go ahead and uh multiply out that binomial. So I'll have X squared plus two A X plus A squared. So if I look at that quantity of MG divided by K, I can actually come back to our equilibrium condition where the weight is equal to the spring force. And so that for X, when I do that I have X is equal to MG divided by K. So I can replace the MG divided by K and our energy conservation equation with X. So we'll do that now. So we'll have two X multiplying the quantity of H plus X plus A. That's gonna be equal to X squared plus two A X plus A squared. So I'll go ahead and multiply through the two X. So I'll have two HX plus two X squared plus two A X and that's gonna be equal to X squared plus two A X plus A squared. Now, a couple of things will cancel out. I have a two A X on both sides of the equations and can subtract those off. I can also subtract one of the X squared terms. And so what I'm left with when I collect everything on the left hand side of the equation, I'll have X squared plus two HX minus A squared equals zero. So here I have a quadratic equation for X. So we can use the quadratic formula to solve it. That means X is going to equal negative two H plus or minus the square root uh 4h squared minus four, multiplied by negative A squared all divided by you. We can simplify this expression. Yeah. So when I, I can factor out a four out of the square root, the square root of four is two and that will cancel out with the two in the denominator. So the twos and the fours were all canceled. And what I'm left with is X is equal to negative H. And here I want to choose the plus sign because I want the X value to be a positive quantity. They'll have a plus, then I'll have the square root of H squared plus A squared. So I have values for the H and the A. So I can actually plug this into my angular frequency formula. So coming back to my angular frequency, I will have the omega is equal to the square root of G divided by, instead of X, here, I'll have the square root of H squared plus A squared. Then I'll have minus H in the denominator. And I have values for the A and H. So I can plug in and calculate my angular frequency. So I have omega equals I'll have the square root four G, that's the acceleration due to gravity. So it's gonna be 9.8 m per second squared and that's gonna be divided by the square root of for H that was five centimeters. So I need to convert that into meters by dividing by 100. So I have 0.05 m and that is squared, they don't have plus the amplitude which was eight centimeters. So that's gonna be 0.08 m that gets squared. And then from that square root, I'll need to subtract ph so that's gonna be the 0.05 m again. So everything on the right hand side is just a numerical value. So I can plug that into my calculator. And if I keep two significant figures, this turns out to be 15 gradient per second. And that corresponds to answer C so just to recap what we did here, we first took our definition for the angular frequency or a mass on a spring. And we had to use the equilibrium condition where the weight is equal to the spring force to figure out how to write the K over M in terms of other quantities that were given. In this case, it turned out to be G divided by X. We then use conservation of energy. You find a relationship between XH and A. And once we did that, we could plug everything back in and calculate our angular frequency. So I hope that this has been useful and I'll see you in the next video.