00:02
Hello, in the question we have given that a set of springs all have initial length 10 centimeters.
00:08
So, here we have given the springs.
00:10
So, it has the initial length of 10 centimeters.
00:14
So, each spring now have mass suspended from its end.
00:20
So, now each spring has a mass suspended at end.
00:26
So, the different spring stretch as shown below.
00:33
So, now when this mass is added, so the spring stretches in accordingly.
00:39
So, now each mass is pulled down additional 1 centimeter and released so that it oscillates up and down.
00:47
So, we have to tell that which system will have the highest frequency.
00:52
So, now the frequency of oscillation is given by, so we know that omega is equal to square root of k by m and omega is equals to 2 pi f.
01:05
So, from here if i want to find out this frequency, so frequency will be equal to omega by 2 pi.
01:12
So, this is nothing but 1 by 2 pi square root of k by m.
01:20
So, this is the frequency.
01:21
But now over here from this relation, so f is equal to 1 upon 2 pi square root of k by m.
01:28
By using this formula, it is not very intuitive that we can explain the answer because here this mass because what they are saying is it is k by m.
01:40
So, spring constant will be same for every system, only the mass is changing.
01:46
So, if mass is, so this relation gives the inverse relation between frequency and mass.
01:53
But if i see over here, i have two identical masses over here, but the frequency of oscillations for this two system is not same.
02:04
So, then we have to use the equation...