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It is known that, when we describe the spring pendulum, we are bound to use the formula $T = 2\pi \sqrt{m/k}$, however, we can go further and set
$\omega = \frac{2\pi}{T}$
I ponder why is this substitution legal, because there is no angular motion in this pendulum!

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  • $\begingroup$ Related: physics.stackexchange.com/q/1018/2451 and links therein. $\endgroup$ – Qmechanic May 10 '16 at 19:12
  • $\begingroup$ Why wouldn't the substitution be legal? You can substitute anything you want. I can substitute $F = 1025\frac m{2\pi k}$ if I want to, although it wouldn't be terribly useful. $\endgroup$ – immibis May 10 '16 at 23:56
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In this equation $\omega$ does not refer to the speed of angular motion, but the frequency of oscillation when measured in angular terms (usually radians/sec, but it can be degrees/sec). Frequency is usually measured in cycles per second (Hertz), but it is sometimes more conveniently measured in angular terms, when it is called angular frequency.

The angle which is being measured here is the phase angle, which describes how far through the cycle the oscillation has gotten, as though the oscillation is moving round in a circle at a constant angular speed $\omega$.

You might well be confused because you cannot see any angular motion in the 'spring' pendulum. But it is even more confusing when dealing with the 'string' pendulum because here the displacement is measured by an angle $\theta$, but this angle is not the same as the phase angle (which is usually called $\phi$, the Greek phi for phase). The angular speed is $\large{\frac{d\theta}{dt}}$, which is normally called $\omega$, but this $\omega=\large{\frac{d\theta}{dt}}$ is not the same as the angular frequency $\omega=\large{\frac{d\phi}{dt}=\frac{2\pi}{T}}$. Angular speed varies during the oscillation, from $0$ at the extremes to the maximum as it passes through the vertical. But angular frequency does not change during the oscillation, it is a constant for ideal string or spring pendulums, depending on $\large{\frac Lg}$ or $\large{\frac km}$.

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Angular velocity

Imagine a point $P$ moving on a circle of radius $R$ with angular velocity $\omega$.

The projection of $P$ onto the $y$-axis is:

$$y=R\sin \theta=R\sin \omega t$$

The point $P'$ is in simple harmonic oscillation (SHO).

For the spring mass system it just so happens that:

$$x=A\sin \omega t$$ where: $$\omega=\sqrt{\frac{k}{m}}$$

So although there is no angular motion in the spring mass SHO, in analogy with the SHO of point $P'$ we still call it the angular velocity.

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The equation for the period $T$ comes by using Newton's second law $F=ma$ to obtain the equation of motion of the spring-mass system

$$-kx =ma \Rightarrow a=-\frac k m$$

where $x$ is the displacement from a fixed point and $a$ is the acceleration.

This equation is of the form $a=-\omega^2 x$ where $\omega$ is a constant of the simple harmonic motion.

It can be shown that for simple harmonic motion $ T = \frac{2 \pi}{\omega}$ which gives $T=2\pi \sqrt{\frac{m}{k}}$.

A possible equation for the displacement of the mass as a function of time $t$ is $x= \sin \omega t$.

Since angles are usually measured in radians and time in seconds the unit of the constant $\omega$ is radian per second.

So in this case the motion of the mass is linear and the reason why the unit of angle, the radian, crops up is because of the sine function being part of the equation for displacement.

Your confusion is probably due to the fact that you have met the same symbol $\omega$ as the angular speed which also has the unit radian per second and the same equation $T=\frac{2\pi}{\omega}$ when dealing with circular motion.

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protected by Qmechanic May 15 '16 at 12:10

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