0:00
Hi there.
00:01
So for this problem we have a 2 kilograms mass.
00:05
So the mass is equal to 2 kilograms and is attached to a spring with a stiffness.
00:13
That means it's the spring's constant is equal to 50 newton's per meter and the mass is this place 1 over 4 a distance that we're going to call and that is 1 over 4 meters to the left of the equilibrium point and given a velocity that is equal to 1 meters 1 meters per second.
00:49
So neglecting damping we need to find the equation of motion of the mass along with the amplitude period and frequency and how long after release does the mass pass through the equilibrium position.
01:06
So with that said, we're gonna let y of d be the rightward displacement in meters at the time t seconds.
01:18
Now then we are to solve the initial value problem that is the following with the values given.
01:25
We're gonna have the mass times the second derivative of this function plus this brings constant, which is 50 times y, and this should be equal to zero.
01:42
Now we are given some values, the initial values are that y, evaluated at zero, is equal to minus 1 over 4, and the velocity, which is the derivative of this function with respect to time, evaluated at 0, is 1 meters per second.
02:06
As the problem states, well, it's minus because it is in the opposite side.
02:11
Now, the auxiliary polynomial to this equation is going to be 2 times x squared plus 50.
02:24
So this, when we set this equal to zero, it's going to have roots that are going to be plus and minus 5 times the complex number i.
02:35
So in that sense, the general homogeneous solution for this problem is going to be that jet of d, it has the form of c1, which is a constant sign of five times the time, plus c2, which is another constant that we need to determine, cosine of five times the time t.
03:04
Now, we also will need the prime version of this equation that is derivative of this function with respect to time.
03:15
So we're going to have that this is 5 times the constant c1, cosine of 5 times the time, and this, because this is the derivative of the same function.
03:26
And in here we're going to have minus c2, well, phi, c2, let's, oh, 5.
03:33
C2, the sign of 5 times the time t.
03:39
And plugging in the time t equals to 0 to evaluate the first condition, we're going to have that.
03:48
Minus 1 over 4 is equal to y evaluated at 0.
03:54
So from there, we obtained that this is equal to c2.
03:59
That is the constant c2, is equal to this.
04:02
And from the other condition evaluated that we know that minus 1 is equal to prime y evaluated at 0.
04:12
From there we obtained that this is 5 times c1.
04:16
So from this, we obtain that c1 is equal to minus 1 over 5.
04:23
And of course, from the other one, we obtain that c2 is minus 1 over 4.
04:29
So those are the constants that we need to plug into the equation, into the solution for this equation.
04:39
So with that said, if y of d is the right war displacement in meters at these seconds, then we obtained that the solution for this equation is that yet y of d is equal to minus 1 over 5 times the sign of 5...