 Let's solve a couple of questions on analyzing graphs of a spring mass system. So for the first one, we have two blocks connected to identical ideal springs are oscillating on a horizontal frictionless surface. The motion of the blocks A and B over time are shown in the graph below and we can see this graph displacement versus time graph of the two blocks A and B. Why are the periods different for blocks A and B? As always pause the video and give this one a try first. Alright, hopefully you have given us a shot. Now we see two plots on one graph. You see displacement time and there is block A, block B. The question is to compare the periods of A and B and let's look at the options. So option A says block A has a longer period because it has more mass. Block B has a longer period because it has more mass. Block A has a longer period because it has a larger amplitude and we can see that block A, the amplitude of the oscillation of block A is 3 cm whereas for block B it's only 1. And the last option is block B has a shorter period because its smaller amplitude means the spring constant is bigger. Now let's try and recall how did we define time period for a spring mass system. Time period was 2 pi into under root of M divided by K. Nowhere in this equation we see amplitude. So amplitude really has no effect on the period of any simple harmonic oscillator. And time period is the time between two consecutive maximas or minimas. So for block A we can see that time this is 4 seconds and for block B, for block B this will be between two consecutive maximas or minimas. So let's look at minimas for block B. This is 2 seconds. Now time period for block A is more, we can see that it's more compared to block B. And if time period is more and these two blocks are connected to ideal springs, that is written over here, these two blocks are connected to ideal identical springs, which means same same springs, identical springs means the same spring, same spring. So that means a spring constant K is the same for both. This is not changing. So if TB is less, TB is less than TA and K is the same. That means mass of B has to be less than mass of A. So block B's mass is less than block A's mass. Only then the time period of A can be more than the time period of B. So out of the options we can say that this is option A. Okay let's look at one more question. Here again we have two identical blocks which are connected to ideal springs. They are oscillating on a frictional surface and we can see the displacement time how they are oscillating. Again the question is why are the periods different for blocks A and B. Now again let's go back to how we defined time period for a spring mass system that was 2 pi into M divided by K. K is the spring constant, M is the mass of the oscillator. And T time period is the time between two consecutive maximas and minimas. Let's see whose time period is more. So for block A this is the time between two consecutive maximas and that is six seconds. That is six seconds. And for block B between two consecutive maximas we can see this much is that is this much the time this is this is four seconds. So block A's time period is more, TA is more than TB. And now we have two identical blocks which means the mass of these two blocks is the same. So M is the same. But they are connected to different springs because it's not written that they are connected to identical springs it's just that they are connected to ideal springs not identical. So spring constant is different. Now if TA is more than TB and mass is the same for both spring constant for B has to be more than spring constant of A. So the spring constant increases, spring constant increases then time period will decrease and time period for B is less. So spring constant for B, so spring constant for the spring connected to block B has to be more than the spring constant for the spring connected to block A. So out of these four options let's look at option A. Block B has a shorter period which is true because it has a smaller amplitude. Not really because amplitude has no relation with, it has no relation with time period at all. It has no relation with the amplitude coming into the equation. So A is wrong. Block B has a shorter period which is true because it has a larger spring constant and that is also true. Option B looks right. Block C, option C talks about amplitude again it's wrong. Block A has a longer period, true because it has a larger spring constant. No that is wrong it has a smaller spring constant. Block B is connected to a spring which has a larger spring constant. That's why the time period for block B is less. So this one is option B. Alright you can try more questions from this exercise in the lesson and if you're watching on YouTube do check out the exercise link which is added in the description.