00:01
Hello students.
00:02
For part a of this problem we want to find the displacement.
00:07
It is given that mass m equal to 19 kilogram and spring constant k equal to 19 kilogram per second square and damping constant c equal to 13 newton second per meter.
00:22
Now the equation of motion can be written as that is m into d square by d.
00:30
Square plus c into dy by d t plus k y equal to zero substituting the values that is 19 into d square y by d t square plus 13 by d y by d t plus 19 y which is equal to zero now the characteristic equation from equation 1 is given by that is 19 m dash square plus 13 m dash plus 19 equal to 0 from this we can find m dash as that is m dash equal to minus 30 plus or minus root of 13 square minus 4 into 19 into 90 divided by 2 into 90 which is equal to minus 30 plus or minus 35 .7 i divided by 38 which is written as minus 0 .34 plus or minus 0 .94 i then the general solution of equation 1 can be written as that is y of t equal to erase to minus 0 .34t into a cost 0 .94t plus b sign 0 .94t where a and b are constant to be determined from initial conditions.
02:08
That is velocity v of t which is equal to d y of want by d t that is equal to 0 .94 e rise to minus 0 .34 t into minus a sign 0 .94 t plus b cause 0 .94 minus 0 .34 plus b cause 0 .94 minus 0 .34 e -res to minus 0 .34 into a cos .94 t plus b sine 0 .94t.
02:56
We have the initial condition that is at a t equal to 0, y of t equal to 1 meter.
03:04
Then we get a equal to 1.
03:12
Then at t equal to 0, v of t equal to 0, that is 0 equal to 0 .94b minus 0 .39a.
03:24
From this we get b equal to 0 .36.
03:29
Then y of t can be written as y of t equal to e rise to minus 0 .34 t into cos .94t plus 0 .94 t plus 0.
03:42
36 .6.
03:44
.94, that is, let a.
03:51
Kose theta equal to 1 and a sine theta equal to 0 .36.
04:01
Then, a square can be written as a square equal to 0 .36 square, that is, plus 1 square, that is, plus 1 square, that is, a equal to 1 .6...