00:01
Hello, in the question we have given that in the system shown in the following figure the numerical value of m, b and k.
00:08
So, mass is given that is equals to 1 kg, b is given that is 2 newton second per meters, then k is given that is the spring constant.
00:22
So, that is 100 newton per meter.
00:26
Now, this x naught is given because it is mentioned that the mass is displaced by 0 .05 meters and then released without initial velocity.
00:40
So, we have to find out the frequency observed and they have mentioned something.
00:45
So, they have given one equation that is omega d is equals to omega n square root of 1 minus this zeta square.
00:58
So, our role will be to find out this omega n and this zeta so that we can plug over here.
01:05
So, now we will write down the equation of motion.
01:12
So, if we write down the equation of motion, so this will be mx double dot plus bx dot plus kx is equal to 0.
01:23
So, now if i divide both the sides, so if i divide this entire equation by m, so i get x double dot plus b by m x dot plus k by m times x is equal to 0.
01:39
Now, we know that omega square.
01:42
So, this quantity is omega squared that is equals to k by m.
01:48
So, from here omega turns out to be square root of k by m.
01:53
We will call this as omega n.
01:55
So, that should be equal to square root of k is 100 divided by mass is 1...