0
$\begingroup$

Why is there no effect in the mass of the bob on the period of the simple pendulum?

I have found out many different explanations. However, I wasn't able to find a scientific explanation for my research proposal.

$\endgroup$
2
  • 10
    $\begingroup$ What research have you done? $\endgroup$ Sep 26 '20 at 11:04
  • 2
    $\begingroup$ Can you show the findings of your research $\endgroup$
    – Buraian
    Sep 26 '20 at 12:47
10
$\begingroup$

TLDR: Restoring force (force of gravity) is dependent on mass, so mass cancels out in $F=ma$.

We know that force due to gravity $F_g=mg$, and so the part of acceleration contributed by the gravitational force, which is $g$, stays the same. The only other force is the tension by the rod/string, which depends linearly on the mass (Net force $T-mg=\frac{mv^2}{r}$, so $T=mg+\frac{mv^2}{r}$), and so if you also apply $F=ma$, the mass divides out. (Note: from now on everything loses dependence on mass, so answer could technically stop here. But OP specifically requested for period, so I'll go on.) If the acceleration stays the same, if the length and the angle stays the same, then the magnitude and direction of acceleration do not change. Therefore, doesn't matter what mass it is, the initial acceleration is the same, and the acceleration changes with time in the same way, as a function $a(t)$. Therefore, if the acceleration at all times is the same, regardless of the mass, the period $T=f(\theta)\cdot2\pi \sqrt{\frac{l}{g}} $, will not depend on the mass.

$\endgroup$
4
  • 1
    $\begingroup$ -1 This answer confuses the principle that F=Gmm/r^2 (and thus that the force acting on the bob is proportional to the mass of the bob) with the principle that a=F/m (and thus that for a force proportional to the mass on the bob, the acceleration is independent of the mass, and thus that the period is also mass-independent. $\endgroup$
    – RLH
    Sep 28 '20 at 2:04
  • $\begingroup$ @RLH We are assuming that the acceleration is constant for the tiny change in $R$, and so there is no need for the formula $F=\frac{GMm}{R^2}$, $F=mg(=ma)$ suffices. $\endgroup$
    – KingLogic
    Sep 28 '20 at 2:29
  • 1
    $\begingroup$ Sure, use $F=mg$ rather than $F=\frac{GMm}{R^2}$. The point is that $F=mg$ is a prescription of what force exists on the system, and $F=ma$ is a description of how accelerations and forces are related to each other. Your opening sentence uses $F=ma$ to assert that the force of gravity increases as mass increases, which is a consequence of $F=mg$, not $F=ma$. $\endgroup$
    – RLH
    Sep 28 '20 at 2:36
  • $\begingroup$ @RLH ok I fixed it. Thanks for the catch! $\endgroup$
    – KingLogic
    Sep 28 '20 at 2:47
13
$\begingroup$

The net torque is only determined by gravity, and the force of gravity is proportional to the mass of the bob, so by Newton's second law we lose the dependency on mass. It's the same reason why all objects have the same acceleration in a gravitational field.

$\endgroup$
1
  • 1
    $\begingroup$ @Mohd Remember to up vote all useful answers and to pick one answer as the selected answer that sufficiently answers your question $\endgroup$ Sep 27 '20 at 4:06
12
$\begingroup$

The simple explanation is that the mass of the bob occurs on both sides of the pendulum's equation of motion, and so cancels out. Suppose the bob has mass $m$ and the pendulum has length $l$. Then when the pendulum makes an angle $\theta$ with the vertical the moment of the bob's weight is $mgl \sin \theta$ so the equation of motion of the pendulum is

$\displaystyle ml^2 \frac {d^2 \theta}{dt^2} = - mgl \sin \theta \\ \displaystyle \Rightarrow \frac {d^2 \theta}{dt^2} = - \frac g l \sin \theta$

and so the motion of the pendulum does not depend on its mass.

$\endgroup$
0
6
$\begingroup$

Let me offer a more conceptual explanation. It's slightly less formal, but doesn't require knowing any equations of motion. For what it's worth, in principle, it would also apply to pendulums swinging at relativistic speeds ;)

Consider two pendulums with equal masses $m$, swinging side by side. The two will have equal periods, so if you set them in motion at the same time, the distance between the masses will remain constant. This means that if you connect the two masses the system will behave exactly the same. But the resulting system is just a pendulum with mass $2m$. By the same token, the period is the same for masses $3m$, $4m$, $5m$ and so on. Inverting the reasoning, the period is also the same for masses $m/2$, $m/3$, $m/4$ and so on. Combining the two arguments, the period is also the same for $q m$ for each rational number $q > 0$. Since rational numbers are dense in the reals (and we believe period depends on mass in a continuous manner) it follows that the period is the same for each mass $m'$.

$\endgroup$
3
  • 1
    $\begingroup$ Nice. The usual "things fall at the same speed" version of this is "If a heavier thing falls faster than a light one, and you tie them together with a thread, then the light one would slow down the heavy one, but the assembly consisting of both would fall faster because it's even heavier...contradiction." That "proof" requires an assumption that falling-speed is monotone as a function of mass. Yours proof here only requires continuity. Not clear which is the more compelling hidden assumption, but I like yours a lot. $\endgroup$
    – John
    Sep 27 '20 at 10:48
  • $\begingroup$ I think this has the potential to be misleading...one could make a very similar argument for a pendulum consisting of a mass on a spring, and conclude that this also has a period independent of mass. You need to know more about the system, i.e. that acceleration due to gravity is mass independent, to make this argument. $\endgroup$
    – DavidH
    Sep 27 '20 at 15:32
  • 1
    $\begingroup$ DavidH - not at all. You just need to know that two pieces of inelastic string right next to each other behave the same way as a single piece of string. This argument does not require us to know anything about how acceleration depends on mass. $\endgroup$ Sep 27 '20 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.