00:01
In this question we are given that a spring with a 7 kg mass and a damping constant 2 can be held straight 0 .5 meters beyond its natural length by a force of 1 .5 newtons.
00:13
Suppose the spring is straighted 1 meters beyond its natural length and then relieved with zero velocity.
00:21
So in the notation of the text we are asked the value of c square minus 4 m k.
00:28
So now here m is 7 kg and damping constant that is k equal to here 2.
00:37
So now c square minus 4 m k.
00:41
So c will be here 7 to the power 2 minus 4 m here is 7 and k is 2 here is 2 here.
00:51
So now that will be 49 minus this is 56.
00:55
So when we solve this we get minus of 7.
00:59
So, value of c squared minus 4m k will be minus 7.
01:03
Now, differential equation for this damping system will be m, d2x divided by d t squared, plus c dx divided by d t, plus kx will be 0.
01:17
So now that will be 7 d2x divided by d t square plus 7 dx divided by d t squared plus 7 dx divided by d t, plus this is 3x equal to 0.
01:28
So now this can be written as 7m square plus 7m plus 3.
01:35
This is our auxiliary equation...