A 2.00-kg, frictionless block is attached to an ideal spring with...

Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

Accepted Answer

Everyone in this problem. A cube of mass. 2.6 kg is fixed to one end of a perfect spring. Spring constant K equals 190 newtons per meter. And is placed on a frictionless surface. Okay? When T equals zero. The spring has its unstrapped length, while the cube has a velocity of 5.4 m per second in the negative direction. Were asked to determine the amplitude and phase angle. Okay? And to express the answer in the form of a position time function. Alright? So if we want to write a position time function, let's go ahead and write that in general and that's going to kind of show us what we need to find. Okay, So our position time function recall is going to be X. Of T is equal to a course, Omega T Plus five. Alright. Sophie is going to be the phase angle. Omega is our angular frequency and a is the amplitude. Okay, So we need to go ahead and find those three things. Now, let's just draw a little sketch of what's going on. Okay? We have our spring. Okay? And it's attached to this block Of 2.6 kg. The spring is unstructured case, what's in its resting position here and when it's at its resting position, we call this position X equals zero. And the velocity or the speed of the block at this point is going in the negative direction. So we're gonna say that's to the right Or sorry? To the left. The positive extraction will take to the right. So the negative direction is to the left with a speed V 5.4 m/s. And then we know our spring constant K is 190 Newtons Per meter. Alright, so we need to find a omega and five. Well, let's recall that omega. The angular frequency OK. Is equal to the square root of K over m. Okay, we know. Okay. And we know em so we can find omega right off the bat. Okay, so this is going to be equal to the square root of newton per meter Divided by 2.6 kg. And this is going to give us an angular frequency of eight . m/s. Whoops. And not meters per second. Sorry, my mistake, radiance per second. We're talking about angular frequency radiance per second. Okay. Always check your units. Alright, so we found omega. Okay, now we need to find A and five. So how can we do those? We actually have some information about a position time point. Okay. That we can use to find a value here. Okay. So what do I mean? Well, we know that when T is zero? This spring is unstrap etched. Okay, so at T equals zero. The spring is unstructured etched. And we've already said that when the spring is unstrap etched, the position is considered X equals zero. Okay, so we have this pair T equals zero and X equals zero. Okay, so we use our equation X at zero is going to be equal to a cosine of omega time zero. So this is going to be zero plus five. Okay and we know that that x value is equal to zero here because it's stretched we can divide by the amplitude and this is going to give us the coast of Phi Is equal to zero. Now when we're talking about our phase angle, okay we want it to be less than or equal to pi So we can calculate this coast if I equals zero, this tells us that five is going to be equal to pi over two. Ok. We want to take the just the smallest value that we can that's going to solve this equation. Alright so let me just go back here. This is our omega value. I'm just gonna put a box around it so we don't lose it. This is our five value pi over to the last thing we have to find is our amplitude. A Okay no let's look at what other information were given. Okay well we're given some information about the velocity or the speed of this block. This cube. When we're at the un stretched position X equals zero. Now when we're at X equals zero. Okay let's recall that this is when we have our maximum velocity. Okay so at X equals zero. We have our maximum velocity and we know that we can write the maximum velocity. It's a oh my God. Okay. So when X equals zero, the velocity is equal to v max which is equal to a soulmate. Okay well we know the speed or the velocity here. We're told in the problem that it's 5.4 meters per second. We're trying to find the amplitude a. And we know omega because we calculated it before the angular frequency is 8.5485 radiance per second. And this is going to give us an amplitude A when we divide 0. leaders. Okay. Alright. So we have all the values we needed. We were able to find the angular frequency rate from the equation using K over M. Okay, we used our position time function at a particular point to find our phase angle phi And then we use the information we knew about the speed of that cube at the unstrapped position. Knowing that that's the maximum speed. To find our ample today. So now that we have those three things we can write our equation. So our position time equation is going to be X. Of T. Is equal to the amplitude A. We're gonna use three significant digits. So 0.632 meters times cosign. And again three significant digits for omega. 8.55 T. Plus the phase angle pi over two. Alright, so this is the position time function for this system. And if we go back up to the top and look at our answer choices, we see that we have answer choice B. Okay. The position time function is X is equal to 0.632 m times. Co sign 8.55 T plus pi over two. Thanks everyone for watching. I hope this video helped see you in the next one.

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