# Effect of magnetic field on time period of a spring pendulum

Let us consider that we have a spring pendulum, and a magnet nearby. Assuming that the spring is being attracted toward the magnet, does the period decrease?

The formula of time period is = 2𝜋√m/k Will the magnetic have the same effect as increasing g?

• How does the attraction change with the position of the pendulum? – probably_someone May 8 at 12:26
• The spring is attracted to the magnet. Any other position than the one shown above, the spring would want to reach equilibrium position. Therefore, I assume that adding a magnet at the bottom decreases the period because the pendulum tries to reach the above equilibrium position shown in the picture – NoLand'sMan May 8 at 12:31
• The formula of the period for a simple pendulum is $T = \frac{1}{2\pi}\sqrt{\ell/g}$. For an oscillating spring, it's $T = \frac{1}{2\pi} \sqrt{m/k}$. – Michael Seifert May 8 at 12:39
• Also: is this actually for a homework assignment? It's an interesting problem, and I can at least give you some hints about how to think about it even if it's a homework problem. But I'd rather not give you a formal solution if you're supposed to figure it out on your own. – Michael Seifert May 8 at 13:04
• @MichaelSeifert Well, to be completely honest, it is not a homework assignment. I was just looking for topics related to pendulums to research on. I must produce a 4000 word journal on my topic, and I thought of this. And since I was confused of the effect of adding a magnet to spring pendulums, I thought I would ask on here. If I manage to get the answer or at least an idea of the answer, and if it can be analysed in 2000 words, I shall research on it. Thanks for your interest! – NoLand'sMan May 8 at 13:37

Non-technical version: the magnet is pulling the mass down, and the magnitude of this force increases as the spring extends. The spring is pulling the mass up, and the magnitude of this force also increases as the spring extends. The effect of the magnet is therefore to decrease the effective stiffness of the spring, since the magnet is working against the spring. The net force on the mass from the spring & magnet combined increases less quickly than the force from the spring alone would; in other words, adding the magnet is basically equivalent to replacing the original spring with a less stiff one. Since the period of a mass on a spring decreases as the spring's stiffness decreases, the period of the oscillations with the magnet present will be lower than without the magnet present.

Technical version: Let $$z$$ be the vertical coordinate of the mass, with positive $$z$$ measured downwards from the unstretched location of the spring. The mass will experience three forces:

• Gravity, with $$F_g = +mg$$;
• The spring force, with $$F_s = - k z$$; and
• A force $$F_m(z)$$ from the magnet. This force will be (presumably) downwards for all $$z$$ values, and will generally be increasing as $$z$$ increases: $$dF_m/dz > 0$$.

Let $$z_0$$ be the coordinate at which the system is in equilibrium. At this point, the net forces on the mass must cancel: $$F_g + F_s + F_m = mg - k z_0 + F_m(z_0)= 0.$$ Let's now imagine displacing the mass a small distance $$\epsilon$$ from equilibrium; in other words, $$z = z_0 + \epsilon$$. Since $$\ddot{z} = \ddot{\epsilon}$$, Newton's second law becomes: \begin{align} m \ddot{\epsilon} &= F_g + F_s + F_m \\ &= mg - k(z_0 + \epsilon) + F_m(z_0 + \epsilon) \\ &\approx mg - k z_0 - k \epsilon + F_m(z_0) + F_m'(z_0) \epsilon \end{align} where we have expanded $$F_m(z_0 + \epsilon)$$ in a Taylor series in the last step. From the equilibrium condition, this then reduces to $$m \ddot{\epsilon} = -(k - F_m'(z_0)) \epsilon$$ and so the frequency of the oscillations is $$\omega = \sqrt{ \frac{k - F_m'(z_0)}{m}} < \sqrt{\frac{k}{m}}.$$ Since the angular frequency of the mass in the absence of the magnet is $$\sqrt{k/m}$$, we conclude that the period of the oscillations increases in the presence of the magnet, and so the period decreases.

(This all assumes that $$F_m'(z_0) < k$$. If this was not the case, then the equilibrium position would be unstable.)

• Thank you so much for your response! If the magnet were to be moved to the side such that the equilibrium position is not completely vertical downwards, then F′m(z0)>k? – NoLand'sMan May 8 at 23:23
• @pranav: That's a good deal more complicated of a situation, since now not all of the forces will be acting along a straight line (in one dimension); and the analysis above would have to be re-done. However, I suspect that the condition $F'_m(z_0) < k$ is largely independent of the position of the magnet. – Michael Seifert May 9 at 11:52