00:01
Everyone in this question we are given that a body is executing s hm with amplitude a and the maximum velocity that is v max is equal to v -r -s.
00:16
Now we have to find that what will be like the speed when it will be like covering a displacement of agibati.
00:24
That is from the main position.
00:26
So what will be it? so how we are going to solve this? we know that like distance is given as level.
00:31
Y by 2 and v max is like this so we know that v max is equal to like omega a using the relation omega root a square minus x square that is equal v is equal to this thing so putting x is equal to zero we will be getting this so this is like equal to vs this is an important thing you have to note it out i will be using this so v a by 2 is like v of a by 2 is equal to omega root a square minus a square by 4, which will be like root 3, that is equal to like root 3 by 2 times of omega and root a square, that is root 3 by 2 omega -a.
01:17
Now putting the value of like omega -a, which is like vs, we will be getting root 3 by 2 b of s.
01:25
So this is like the speed at displacement of a by 2.
01:31
So how i am saying this thing, that suppose this is like a s -hm doing body, suppose this is like a pendulum.
01:39
So this is like a zero position...