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So I wanted to find out how to (simply, if that's possible) derive the formula for a period of spring pendulum: $T=2\pi \sqrt{\frac{m}{k}}$. However, Google doesn't help me here as all I see is the ready-to-bake formula. Could you please point me some directions?

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Related: physics.stackexchange.com/q/1018/2451 –  Qmechanic Jan 20 '13 at 17:33

3 Answers 3

up vote 2 down vote accepted

You need to know the equation of motion. The force for the pendulum is given by $F= k x$. Newtons equation tell you $F=ma = m \ddot x$. So you need to solve $$\tag{1} m \ddot x = k x.$$

You know that the solution will be of oscillatory form. So you set $x= A \cos(2\pi t/T)$ and you want to obtain $T$. Plugging this ansatz into the equation (1), you obtain $$ m\frac{(2\pi)^2}{T^2} A \cos(2\pi t/T) = k A \cos(2\pi t/T). $$ You see that the equation is fulfilled if $$ m\frac{(2\pi)^2}{T^2} = k.$$ Solving for $T$, you obtain the result.

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I see. Starts to make sense, thank you! :) –  Straightfw Jan 20 '13 at 17:42
@VladimirKalitvianski: thanks for spotting the typo. –  Fabian Jan 20 '13 at 18:17

There's one more simple method for deriving the time period (an add-up to Fabian's answer). The spring pendulum, as we all know is a great (well-known) example for Simple Harmonic Motion. First, let's assume a particle at any point of the spring. While in an SHM, the acceleration of the particle (at some position on the pendulum) is directly proportional to its displacement ($x$), and is always directed towards the fixed point of the spring. From the past observations, $$a\ \alpha\ x \implies a=-\omega^2x$$

I think you know that $\omega$ is the angular frequency of this particle and the negative sign indicates that the acceleration is opposite to the direction of displacement. If the mass is given by $m$, then the restoring force is given by substituting the above equation in Newton's second law. (For instance, it is $F=ma$)

The restoring force, $$F=-m\omega^2x$$ Here, the $m\omega^2$ is substituted by $k$. Its always a constant for a given spring (and hence the name, spring constant). From this, we could simply get $$\omega=\sqrt\frac{k}{m}$$

Now, time period $T$ is the time taken by the spring to complete one oscillation with a given $\omega$. We could define angular velocity using this time period. A complete revolution per $T$, gives the angular frequency. $$\omega=\frac{2\pi}{T}$$

Equating both, you obtain $$T=2\pi \sqrt{\frac{m}{k}}$$

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OK, thank you a lot :) –  Straightfw Jan 20 '13 at 18:41

It is also possible to dimensionally estimate the period from the equation of motion without solving it, if you roughly replace the second time derivative $\ddot x$ with $L/(T/2)^2$:

$$m\frac{L}{(T/2)^2}=kL,\;\to T\propto 2\sqrt{\frac{m}{k}}.$$ You see, $T/4$ is even better characteristic time estimation for the time derivative.

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I see. Thank you too! :) –  Straightfw Jan 20 '13 at 18:43

protected by Qmechanic Aug 19 at 13:57

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