00:01
In this problem, first i'm just writing the given data here.
00:05
So here the value of mass of instrument is given by m is equal to 20 kg.
00:11
Now i can write the value of omega 1 is equal to 2 pi n1 by 60 which is equal to 2 pi multiplication 1000 by 60 which is equal to 104 .719 radian per second.
00:29
The value of omega 2 is equal to 2 pi n2 by 60, which is equal to 2 pi multiplication 3 ,000 by 60, which is equal to 314 .159 radian per second.
00:46
Now in the next step, i can write the natural frequency formula omega n is equal to under root k by m, which is equal to under root k by 20.
00:59
Which is equal to 0 .226 multiplication under root k.
01:07
Now the range ratio is given by r1 is equal to omega 1 by omega n which is equal to 104 .719 by by 0 .2 .36 multiplication root k.
01:24
The value of r2 is equal to omega 2 by omega n.
01:29
Omega n which is equal to 314 .159 by 0 .22 .26 root k.
01:45
So it will be equal to so it will be equal to 1405 .0044 by root k.
01:58
Now i can write the value of td is equal to x by which is equal to 0 .1...