WEBVTT
00:00:00.230 --> 00:00:04.030
The graph shows the length of a spring as the force applied to it changes.
00:00:04.400 --> 00:00:05.960
What is the spring constant?
00:00:06.500 --> 00:00:13.800
Okay, so as we can see here, in the graph, we’ve been given the length of the string on the horizontal axis and the force applied to the spring on the vertical axis.
00:00:14.350 --> 00:00:21.840
Now, we can clearly see that when no force — zero force — is applied to the spring, the length of the spring is 0.5 meters.
00:00:22.310 --> 00:00:25.490
Therefore, this is what the spring looks like when no forces have been applied to it.
00:00:25.880 --> 00:00:28.770
So here’s the spring and its length is 0.5 meters.
00:00:29.390 --> 00:00:32.340
Now, subsequently, the force applied to the spring increases.
00:00:32.770 --> 00:00:36.360
And we can see that as the force applied increases, so does the length of the spring.
00:00:36.950 --> 00:00:42.230
In other words, we exert a force on the spring — let’s say on the right end of the spring — and we call this force 𝐹.
00:00:42.770 --> 00:00:45.360
Now because of this, the length of the spring increases.
00:00:45.680 --> 00:00:49.760
So now, the total length of the spring is from here to here.
00:00:50.480 --> 00:00:54.710
Now, if we’ve been asked to find the spring constant of the spring, then we need to recall Hooke’s law.
00:00:55.130 --> 00:01:03.270
Hooke’s law tells us that the force applied to a spring 𝐹 is equal to the spring constant of the spring 𝑘 multiplied by the extension of the spring 𝑥.
00:01:03.810 --> 00:01:07.790
Now, note that we’re talking about the extension of the spring and not the length of the spring.
00:01:08.390 --> 00:01:15.530
In other words, when the force 𝐹 has been applied to the spring, this force is directly proportional to the extension; that’s this length here.
00:01:15.980 --> 00:01:17.610
So let’s call that length 𝑥.
00:01:18.140 --> 00:01:22.350
The force is not directly proportional to the entire length of the spring, which we’ll call 𝐿.
00:01:22.900 --> 00:01:32.430
But because we’ve been given the length of the spring on the horizontal axis, we need to find a relationship between the length of the spring and the extension so that we can work out the spring constant.
00:01:32.790 --> 00:01:45.400
The way we do this is to say that the total length of the spring — this whole distance here — is equal to 0.5 — that’s the natural length of the spring when no force is acting on it — plus 𝑥, the extension.
00:01:45.840 --> 00:01:55.090
And so we can say that 𝐿 is equal to 𝑥 plus 0.5 meters or if we rearrange this, 𝑥 is equal to 𝐿 minus 0.5 meters.
00:01:55.530 --> 00:02:03.510
This means that for any point on the graph, we can now work out the force exerted on the spring, which we can simply read off by moving left to the vertical axis.
00:02:03.840 --> 00:02:05.830
And we can work out the extension of the spring.
00:02:06.210 --> 00:02:13.230
Because we know the length of the spring at that point and if we subtract 0.5 meters from it, then we’ll have the extension.
00:02:13.840 --> 00:02:19.580
So let’s just pick any random point and rearrange Hooke’s law to work out the spring constant.
00:02:20.070 --> 00:02:25.970
We’re gonna divide both sides of the equation by 𝑥, the extension, so that the extension cancels out on the right-hand side.
00:02:26.460 --> 00:02:32.560
What we’re then left with is that the spring constant 𝑘 is equal to the force applied 𝐹 divided by the extension 𝑥.
00:02:32.970 --> 00:02:35.280
And now, let’s assume we’re considering this point here.
00:02:35.830 --> 00:02:41.630
Well, we can see that the force applied at this point, if we go left to the vertical axis, is 200 newtons.
00:02:42.270 --> 00:02:51.820
So we can say that when the force 𝐹 is 200 newtons, the length of the spring 𝐿 is going to be 3.0 meters, which is what we write down over here.
00:02:52.320 --> 00:02:55.580
But then, we can use this equation to give us the extension of the spring.
00:02:56.060 --> 00:02:58.850
Essentially, we subtract 0.5 meters from the length 𝐿.
00:02:59.270 --> 00:03:06.590
And so we say that 𝑥, the extension, is equal to 3.0 meters, the length, minus 0.5 meters, the natural length of the spring.
00:03:06.990 --> 00:03:10.620
And when we evaluate this, we find that the extension is 2.5 meters.
00:03:11.060 --> 00:03:17.390
So now, we’ve got the force applied to the spring on one point of the graph and the extension of the spring at that same point.
00:03:18.020 --> 00:03:20.920
Therefore, we can substitute in the values to find the value of 𝑘.
00:03:21.370 --> 00:03:34.370
Now, note that what we’re basically doing is finding the gradient of this line of best fit because what we’re doing here is saying that 𝑘 is equal to the force applied to the spring which is 200 newtons.
00:03:34.370 --> 00:03:38.180
But that’s also equivalent to 200 minus zero newtons.
00:03:38.430 --> 00:03:45.340
And we’re dividing this by 3.0 — that’s the length of the spring — minus 0.5 — that’s the natural length of the spring.
00:03:45.730 --> 00:03:50.350
So coincidentally, by doing this calculation, we’re actually finding the gradient of the line of best fit.
00:03:50.770 --> 00:03:52.020
But this is irrelevant.
00:03:52.230 --> 00:04:00.030
Now, when we sub in the values, we can say that the value of 𝑘 is equal to the force 200 newtons divided by the extension 2.5 meters.
00:04:00.490 --> 00:04:05.980
And once we evaluate the fraction, on the right, we find that the value of 𝑘 is 80 newtons per meter.
00:04:06.500 --> 00:04:07.890
Hence, we have a final answer.
00:04:08.300 --> 00:04:11.380
The spring constant of the spring is 80 newtons per meter.