(III) A mass m is at rest on the end of a spring of s...

Question

(III) A mass m is at rest on the end of a spring of spring constant k. At t = 0 it is given an impulse J by a hammer. Write the formula for the subsequent motion in terms of m, k, J, and t.

Accepted Answer

Welcome back. Everyone. In this problem. A block of mass M is attached to the free end of a horizontal spring. The other end of the spring is fixed to a wall at T equals zero. A child kicks the block with an impulse J given that the spring constant is C determine the expression describing the resulting motion of the block. In terms of JMC. And T assume that there is no friction involved for our answer choices. A says that motion X is J divided by the square root of CM multiplied by the cosine of the square root of C divided by M multiplied by T B says it's J divided by the square root of CM multiplied by the sine of the square root of C divided by M multiplied by TC says it's J multiplied by the square root of C divided by M cubed multiplied by the cosine of the square root of M divided by C multiplied by T and D says it's J multiplied by the square root of C divided by M cubed multiplied by the sine of the square root of M divided by C multiplied by T. Now, if we want to first figure out the resulting motion of the block in terms of these variables that we know, let's ask ourselves, what do we know about the motion of a block oscillating in simple harmonic motion will recall that our motion X is going to be equal to a multiplied by some trigonometric function multiplied by omega T. Now, in this case, that trigonometric function is going to be the sine function. And that's because when T equals zero, it tells us that that's where the child kicks the block and thus X equals zero. So since X and T equals zero at the same time to start, then it's going to be a sine function. A sine graph. OK. In other words, it would look something like this. So we know what type of trick function it is. But no, we just need to find expressions for the amplitude and our frequency omega in terms in terms of our uh our, our variables here, JMC and T. So what do we know that can help us? Well, first, let's start with our angular frequency. Now we know that our angular frequency Omega is equal to the square root of the spring constant K divided by the mass. So in this case, our spring constant is C OK, our mass is M. So Omega equals the square root of C divided by M. Next, let's try to find the amplitude. What can help us to find the amplitude Well, so far we know that the child kicks the block with an impulse of chase. I'm just gonna underline this here to show that we know what our angular frequency is. And since our child kicks the block with an impulse J, what do we know about impulse? Well, OK. Say it on here we're solving for a recall that impulse J OK is equal to the change in momentum. OK? No, our momentum is equal to mass multiplied by velocity. So in this case, that change in momentum equals M multiplied by our change in velocity. Remember that for this impulse, it causes the block at rest to move at an initial speed V knot. So we can replace or change in speed as V knot minus zero from rest which when we expand, this is going to be equal M multiplied by V knot. Now, since V knot is the speed at the equilibrium, it is the maximum speed of the block. In other words, V knot is also equal to the max. Now, since here we know that J is equal to M multiplied by V max solving for V max, OK. This means that V max is going to be equal to our impulse J divided by our mass M. Now, you might be wondering what does the maximum speed have to do with the amplitude? We'll recall that the more the more vibrations there are, the faster the object object gets. In other words, the maximum speed is directly proportional to the angular frequency and the amplitude. And we know that because OK, because let me write it below it here, we know that because our maximum speed V max, OK is equal to our angular frequency multiplied by our amplitude. We already have an expression for our angular frequency. We know that it's the square root of C divided by M. So that is going to be equal to the square root of C divided by M multiplied by or amplitude A. And thus solving for our amplitude, OK, if we equate both sides because remember V max equals J divided by M, then that means J divided by M is going to be equal to a multiplied by the square root of C divided by M. Now here, if we go ahead and solve for A, then that means A is going to be equal to J divided by M multiplied by the square root of C divided by M. Now let's just do a little math here. Uh We know that we can rewrite this as J divided by MK divided by the square of M multiplied by C OK. And if we think about it, we could rewrite M as the square root of M squared. OK? Because the square would cancel the square root. And now when we put one term into another, OK? Or this should be square under the square root as well. That means J equals the square root of C multiplied by the square root of M which we can rewrite as the square root of CM based on what we know about roots. So here we have this expression for the amplitude of our system. So now that we have an expression for our amplitude and we already found one for our angular frequency. If we put that into our general equation, OK, then there are four. Yeah, that means OK, that A is going to be equal to AJ divided by the square root of cm multiplied by the sine of Omega which is the square root of C divided by MT. Therefore, if we look back at our answer choices that tells us, then that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

(III) A mass m is at rest on the end of a spring of spring constant k. At t = 0 it is given an impulse J by a hammer. Write the formula for the subsequent motion in terms of m, k, J, and t.

Key Concepts

Impulse

Spring Constant

Simple Harmonic Motion

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