00:01
Hello students, in this question we have provided with a spring mass damper system, which means it connect masses connected by a spring and dampers used to simulate oscillatory movement.
00:16
So, depending on the external forces.
00:18
So, given that k is equal to spring constant is 4000 newton per meter, mass is 10 kg, then damping coefficient c is equal to 40 newton second per meter, then given equation of force f of t is equal to 200 cos 10 t.
00:52
So, now this comparing with a cos omega t, we can write f 0, initial f 0 is equal to 200 newton and omega angular velocity is 10 radian per second.
01:17
So, first we have to look at the natural frequency say omega 0, which is equal to k by m under root.
01:30
So, which is equal to 4000 divided by 10 all raised to 1 by 2, which is equal to 20 radian per second, this is the natural frequency.
01:46
So, we can look of the static deflection which gives.
01:49
So, delta static is equal to f 0 by k force by spring constant.
02:01
So, which is equal to 200 newton divided by 4000 newton per meter, which is equal to 0 .05 meter.
02:17
Thus, we can calculate the damping factor tau is equal to c divided by 2 into k m all raised to 1 by 2, which is equal to damping coefficient, 40 divided by 2 into 4000 into 10 all raised to 1 by 2.
02:42
Now, we can look at the damping dynamic frequency.
02:46
So, for looking dynamic frequency, we can write omega d is equal to omega 0 into 1 minus tau square all raised to 1 by 2, which is equal to 20 into 1 minus 0 .1 all square all raised to 1 by 2, which is equal to 19 .8 radian per second.
03:21
So, first then we can find the ratio of omega by omega 0, which is equal to 10 by 20, which is equal to 0 .5.
03:35
Thus, we can look at the displacement.
03:39
So, x is equal to delta static divided by 1 minus the ratio r square all square plus 2 into damping coefficient 2 tau r all raised to all square, which is raised to 1 by 2...