As an example of basic harmonic motion, we very first consider the activity of a block of massive (m) that have the right to slide there is no friction follow me a horizontal surface. The mass is attached to a spring v spring continuous (k) which is attached come a wall on the other end. We present a one-dimensional coordinate system to define the position of the mass, such that the (x) axis is co-linear through the motion, the beginning is located where the feather is at rest, and the confident direction synchronizes to the feather being extended. This “spring-mass system” is shown in figure (PageIndex1).

You are watching: Suppose you have a mass m attached to a spring Figure (PageIndex1): A horizontal spring-mass system oscillating about the beginning with one amplitude (A).

We assume that the pressure exerted by the feather on the fixed is offered by Hooke’s Law: <eginaligned vec F = -kx hat xendaligned> where (x) is the position of the mass. The only other forces exerted on the mass space its weight and the normal force from the horizontal surface, which are equal in magnitude and also opposite in direction. Therefore, the net pressure on the mass is the force from the spring.

As we observed in section 8.4, if the feather is compressed (or extended) through a distance (A) loved one to the remainder position, and the massive is then released, the mass will certainly oscillate ago and forth between (x=pm A)1, which is illustrated in figure (PageIndex1). We contact (A) the “amplitude the the motion”. Once the massive is in ~ (x=pm A), its rate is zero, together these points correspond to the location where the mass “turns around”.

Exercise (PageIndex2)

What is the SI unit for angular frequency?

( extHz) ( extrad/s) ( extN^1/2 extm^-1/2 extkg^-1/2) every one of the over Answer

All that the above

Olivia"s Thoughts

In Chapter3, us found, (x(t)), from a function, (a(t)), through using an easy integration. You may be wondering why we can’t execute the same thing in order to find (x(t)) for the mass-spring system. The difference is that, before, the acceleration was a role of time. Here, the acceleration is a duty of (x). This way that we need to use a different an approach to deal with for (x(t)), i beg your pardon is why we are making this “guesses” to resolve a differential equation.

We still require to identify what the constants (A) and (phi) have to do through the movement of the mass. The continuous (A) is the maximal value that (x(t)) deserve to take (when the cosine is equal to 1). This coincides to the amplitude that the motion of the mass, i beg your pardon we currently had labeled, (A). The constant, (phi), is referred to as the “phase” and depends on as soon as we choose (t=0) come be. Intend that we define time (t=0) come be as soon as the massive is at (x=A); in the case: <eginaligned x(t=0) &= A\ A cos(omega t + phi) &= A\ A cos(omega (0) + phi) &= A\ cos(phi) &= 1\ herefore phi = 0endaligned> If we specify (t=0) come be once the mass is in ~ (x=A), climate the phase, (phi), is zero. In general, the worth of (phi) have the right to take any value between (-pi) and also (+pi)3 and, juniorg8.comically, corresponds to our choice of once (t=0) (i.e. The place of the mass as soon as we select (t=0)).

See more: Wide And Narrow Vs Wide And Narrow Type, Difference Between Wide Feet And Narrow Feet

Since we have figured out the place as a duty of time because that the mass, the velocity and also acceleration together a function of time are easily discovered by acquisition the corresponding time derivatives: <eginaligned x(t) &= A cos(omega t + phi)\ v(t) &= fracddtx(t) = -Aomegasin(omega t + phi)\ a(t)&= fracddtv(t) = -Aomega^2cos(omega t + phi)endaligned>

Exercise (PageIndex3)

What is the worth of (phi) if we pick (t=0) to be when the fixed is in ~ (x=0) and also moving in the confident (x) direction? 