A mass of 042 kg is attached to a spring and set into...

A mass of 0.42 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.40 m)cos[(10 rad/s)t]. Determine the following. (a) Amplitude of oscillation for the oscillating mass (m) (b) Force constant for the spring (N/m) (c) Position of the mass after it has been oscillating for one half a period (m) (d) Position of the mass one-third of a period after it has been released (m) (e) Time it takes the mass to get to the position x = -0.10 m after it has been released (s)

Transcript

00:01 So in this question, we are to answer various questions concerning an object undergoing simple harmonic motion.

00:09 And raven told the formina for that motion is x is equal to 0 .40 meters times the cosine of 10 radians per second times the time.

00:34 The first thing we are asked is what is the amplitude of this motion? well, just from inspection of the above equation, the amplitude is the number that you multiply times the trigonometric function there is going to be this number in front.

00:52 And so the amplitude here is going to be 0 .40 meters.

01:00 That means the biggest that this displacement can be is 0 .4 meters.

01:05 Or negative 0 .4 meters because the biggest the cosine can be is 1 or negative 1, i suppose, on the other side.

01:14 The next question, either you already know the formula for the, that relates the angular frequency to the mass and the spring constant.

01:29 That formula is that omega squared, where omega is this angular frequency here, 10 radians per second.

01:36 Is equal to k, the spring constant or the force constant, divided by the mass that is oscillating.

01:45 To see this, if you have a formative chart that just gives you, say, for example, the period, well, the period, it would be given then as 2 pi times the square root of m over k.

02:01 Well, the period is one over the frequency, so the frequency that, then would be 1 over 2 pi times the square root of k over m.

02:15 And then omega, which was equal to 2 pi times the frequency, would be equal to the square root of k over m.

02:27 And if you square both sides of that equation, you get the equation that i came up with earlier.

02:33 So in this case, then, we're trying to find k.

02:35 So what we're going to say is that m times omega squared is equal to k.

02:48 And we are told here that the mass is 0 .42 kilograms and omega is 10.

02:58 So omega squared would be 100.

03:03 And that would be radians squared over seconds squared.

03:15 And what we end up with is that the k is equal to 100 times 0 .42.

03:23 So that's going to be equal to 42 newtons per meter.

03:29 That would be the unit that we would find here...