As an example of an easy harmonic motion, we an initial consider the motion of a block of mass $$m$$ that have the right to slide without friction along a horizontal surface. The mass is attached to a spring through spring constant $$k$$ i beg your pardon is attached to a wall on the various other end. We present a one-dimensional coordinate system to explain the place of the mass, such that the $$x$$ axis is co-linear with the motion, the beginning is located where the feather is at rest, and also the hopeful direction synchronizes to the feather being extended. This “spring-mass system” is illustrated in number $$\PageIndex1$$.

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Figure $$\PageIndex1$$: A horizontal spring-mass device oscillating about the beginning with one amplitude $$A$$.

We assume the the force exerted by the feather on the fixed is offered by Hooke’s Law: \<\beginaligned \vec F = -kx \hat x\endaligned\> wherein $$x$$ is the place of the mass. The just other forces exerted ~ above the mass space its weight and the normal force from the horizontal surface, which space equal in magnitude and also opposite in direction. Therefore, the net force on the mass is the pressure from the spring.

As we saw in ar 8.4, if the spring is compressed (or extended) by a street $$A$$ loved one to the remainder position, and also the mass is climate released, the mass will certainly oscillate earlier and forth in between $$x=\pm A$$1, i m sorry is depicted in figure $$\PageIndex1$$. We contact $$A$$ the “amplitude that the motion”. When the fixed is in ~ $$x=\pm A$$, its speed is zero, as these points correspond to the ar where the massive “turns around”.

Exercise $$\PageIndex2$$

What is the SI unit because that angular frequency?

$$\textHz$$ $$\textrad/s$$ $$\textN^1/2\textm^-1/2\textkg^-1/2$$ all of the above Answer

All the the above

Olivia"s Thoughts

In Chapter3, us found, $$x(t)$$, indigenous a function, $$a(t$$), by using an easy integration. You may be wonder why us can’t execute the very same thing in order to discover $$x(t)$$ for the mass-spring system. The distinction is that, before, the acceleration was a duty of time. Here, the acceleration is a duty of $$x$$. This method that we have to use a different an approach to settle for $$x(t)$$, which is why we are making these “guesses” to resolve a differential equation.

We still need to identify what the constants $$A$$ and also $$\phi$$ have to do through the motion of the mass. The constant $$A$$ is the maximal value that $$x(t)$$ can take (when the cosine is same to 1). This synchronizes to the amplitude the the activity of the mass, i m sorry we currently had labeled, $$A$$. The constant, $$\phi$$, is referred to as the “phase” and also depends on once we choose $$t=0$$ to be. Suppose that we define time $$t=0$$ to be when the massive is in ~ $$x=A$$; in that case: \<\beginaligned x(t=0) &= A\\ A \cos(\omega t + \phi) &= A\\ A \cos(\omega (0) + \phi) &= A\\ \cos(\phi) &= 1\\ \therefore \phi = 0\endaligned\> If we specify $$t=0$$ to be when the mass is in ~ $$x=A$$, climate the phase, $$\phi$$, is zero. In general, the worth of $$\phi$$ have the right to take any type of value between $$-\pi$$ and $$+\pi$$3 and, jonathanlewisforcongress.comically, synchronizes to our an option of once $$t=0$$ (i.e. The position of the mass as soon as we select $$t=0$$).

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Since us have figured out the position as a duty of time for the mass, that is velocity and also acceleration together a duty of time are easily discovered by taking the equivalent time derivatives: \<\beginaligned x(t) &= A \cos(\omega t + \phi)\\ v(t) &= \fracddtx(t) = -A\omega\sin(\omega t + \phi)\\ a(t)&= \fracddtv(t) = -A\omega^2\cos(\omega t + \phi)\endaligned\>

Exercise $$\PageIndex3$$

What is the worth of $$\phi$$ if we pick $$t=0$$ come be once the fixed is in ~ $$x=0$$ and moving in the confident $$x$$ direction?

$$\pi$$ $$-\pi$$ $$\pi/2$$ $$-\pi/2$$ Answer

The position of the fixed is described by a sinusoidal role of time; we contact this type of motion “simple harmonic motion”. The position and also velocity as a function of time for a spring-mass device with $$m=1\textkg$$, $$k=4\textN/m$$, $$A=10\textm$$ are displayed in number $$\PageIndex2$$ for 2 different selections of the phase, $$\phi=0$$ and $$\phi=\pi/2$$.

Olivia"s Thoughts

Here’s a visualization of uniform circular activity projected onto the $$x$$ axis: