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I have been researching for almost a week looking for some variable that might affect the time period in a pendulum other than length and the only thing I found is the medium and that is due to resistivity. Does anyone have an idea of a variable that might have an effect on the time period of a pendulum?

Thanks a lot.

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The initial condition affects the amplitude. From a free body diagram and using that the power is null (energy concervation $\frac{\mathrm{d} E_{mec}}{\mathrm{d} t} = 0$ ) you find as solution (forgetting the trivial solution $\dot\theta = 0$, where $\theta$ is the angle of the pendulum)

$$ \ddot\theta = -\omega_0 \sin\theta ,$$

with $\omega_0^2= \frac{g}{l}$.

Then expanding the $\sin$ to first order (linear approximation) you'd find

$$\ddot\theta = \theta$$ which is an harmonic oscillator and has solution

$$\theta(t) = \theta_0\sin{\left(\omega_0 t\right)}$$.

With $\theta_0$ being the drop angle. Now you can already see that the period depends on the gravity and length (so you already found an other one, i.e. gravity).

But one can do more. The $\sin$ approximation is getting valid only when the angle gets small (how small?). One could get a better approximation by expanding to higher orders. $\sin x = x - \frac{x^3}{6} + o(3)$. Then one would look for solutions of the differential equation

$$\ddot \theta = \omega_0^2\left(\theta - \frac{1}{6}\theta^3\right)$$

sometimes called the Duffing equation. It requires advanced tools to be solved (such as Fourier series expansion). I can tell you that the solution would have a period

$$T = \frac{2\pi}{\omega_0}\left(1+\frac{\theta_0^2}{16}\right)$$.

So you see that with this better approximation the period of the pendulum depends on the drop off angle. Which intuitively makes sense.

If you don't want to solve the equations, you could also keep the $\sin$ formulation and solve the system numerically (implicit Euler or Runge-Kutta for instance) and you would see that the oscillations depend on the drop off angle. Or check by yourself with a pendulum (lab style).

If you have friction the energy will not be conserved and the period will not even be a constant of motion. The period would actually decrease in time.

I hope this helps :), Best

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  • $\begingroup$ Thanks a lot @kyril for your answer. $\endgroup$ – user276286 Nov 11 '20 at 13:51
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Assuming you are talking about the simple pendulum and concern with minor effects too.

First of all the time period of the pendulum depend on Amplitude. Because for large amplitudes the approximation $\sin\theta\approx \theta$ may not work very well. $$T=2\pi\sqrt{\frac{l}{g}}\left[1+\frac{1}{16}\theta_0^2+\cdots\right]$$

  • Temperature : The largest source of error in early pendulums was slight changes in length due to thermal expansion and contraction of the pendulum rod with changes in ambient temperature.
  • Atmospheric pressure : The effect of the surrounding air on a moving pendulum is complex and requires fluid mechanics to calculate precisely, but for most purposes its influence on the period can be accounted for by three effects.

By Archimedes' principle the effective weight of the bob is reduced by the buoyancy of the air it displaces, while the mass (inertia) remains the same, reducing the pendulum's acceleration during its swing and increasing the period. This depends on the air pressure and the density of the pendulum, but not its shape.

The pendulum carries an amount of air with it as it swings, and the mass of this air increases the inertia of the pendulum, again reducing the acceleration and increasing the period. This depends on both its density and shape.

Viscous air resistance slows the pendulum's velocity. This has a negligible effect on the period, but dissipates energy, reducing the amplitude. This reduces the pendulum's Q factor, requiring a stronger drive force from the clock's mechanism to keep it moving, which causes increased disturbance to the period.

Gravity : Pendulums are affected by changes in gravitational acceleration, which varies by as much as $0.5$ % at different locations on Earth, so precision pendulum clocks have to be recalibrated after a move. Even moving a pendulum clock to the top of a tall building can cause it to lose measurable time from the reduction in gravity.


Source : Wikipedia

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  • $\begingroup$ Thanks a lot @Young Kindaichi for your valuable answer. $\endgroup$ – user276286 Nov 11 '20 at 13:52
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The equation of motion of an ideal pendulum in vacuum is $$ \ddot \theta + \frac g l \sin \theta=0 $$ where $\theta$ is the angle the pendulum makes with the vertical, $g$ is the gravitational acceleration and $l$ is the length. For small $\theta$, the equation can be approximated to $$\ddot \theta + \frac g l \theta=0$$ which describes simple harmonic motion with period $$T = 2\pi \sqrt{\frac l g}. $$ This means that for small oscillation amplitudes, the period depends on the length and the gravitational acceleration. Accelerating the "ceiling" the pendulum is attached to (think an elevator) upward or downward would also be equivalent to a change in $g$ and affect the period.

In the original equation without the small angle approximation, the second (restoring force) term is proportional to $\sin \theta$, which is actually smaller than the $\theta$ factor in the harmonic oscillator restoring force. Since the restoring force is smaller compared to the harmonic oscillator, the pendulum is pulled toward $\theta = 0$ with less acceleration compared to a harmonic oscillator, so the period is larger. Thus, the oscillation amplitude also affects the oscillation period: the greater the amplitude, the worse the small angle approximation and the longer the period.

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  • $\begingroup$ Thanks a lot @Puk for your answer. $\endgroup$ – user276286 Nov 11 '20 at 13:52

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