# Why doesn't mass of bob affect time period?

The gravitation formula says $$F = \frac{G m_1 m_2}{r^2} \, ,$$ so if the mass of a bob increases then the torque on it should also increase because the force increased. So, it should go faster and thus the oscillation period should be decrease.

My physics book says that period is only affected by effective length and $g$. Why doesn't mass of bob affect the period?

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The question is not clear to me. Can you explain the situation a bit more? – Ali Jul 21 '13 at 13:59
ok,let me make the question clearer – svineet Jul 21 '13 at 14:05
Hint: Why doesn't the mass of an object affect how long it takes to fall from a given height to the ground? Mass doesn't only appear in the gravitation formula: It also appears in $\vec F=m\vec A$ – james large Sep 14 at 21:28
You're hitting on the Equivalence principle and the Eötvös experiment. – WetSavannaAnimal aka Rod Vance Sep 14 at 22:54

A very loose answer would be that the time period actually depends upon the angular acceleration and not the torque.

Just like the time taken for a object to fall through a height of $h$, depends on the gravitational acceleration and not the mass, i.e. if you drop a sponge ball or you jump yourself, you both will cover height $h$ in the same time(of course neglecting air resistance).

Similarly, the time period of a pendulum doesn't depend upon the mass, or rather the inertia of the pendulum, but only on the angular acceleration due to gravity.

Now you might ask that in this case, it should also not depend upon the length, but the term of length comes when you calculate the angular acceleration due to the acceleration of gravity.

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For the same reason objects of different masses fall at the same acceleration (neglecting drag): because while the force is proportional to the mass and the acceleration is inversely proportional to mass.

Doing the falling case o avoid having to deal with the vectors in the pendulum we get

$$a = \frac{F}{m} = \frac{G\frac{Mm}{r^2}}{m} = G\frac{M}{r^2}$$

where $M$ is the mass of the planet, $m$ is the mass of the object you are dropping and $r$ is the radius of the planet.

The mass of the minor object falls out of the kinematics.

The same thing happens in the case of the pendulum: the force includes a factor of $m$, but the acceleration does not.

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A pendulum in a gravitational field experiences an instantaneous torque about its pivot point of $$\vec{\Gamma} = \vec{r}\times m\vec{g}$$ where $\vec{g}$ is the instantaneous gravitational field, and $r$ is the distance from the pivot point to the CoM.

For purposes of this answer $$\vec{g}=-G\frac{ M_E}{(R_E + h)^2}\hat{k}$$ where

• $m$ is the pendulum mass,
• $M_E$ is the Earth's mass,
• $R_E$ is the distance from the gravitational center of the Earth to the center of mass of the pendulum at rest, and
• $h=r(1-\cos\theta)$ is the height of the CoM when the pendulum is oscillating.

Let's assume the pendulum is oscillating in a plane, so we can write $$\Gamma = mgr\sin\theta = \mathcal{I}\frac{d^2\theta}{dt^2}.$$

$\mathcal{I}$ is the moment of inertia of the pendulum about the pivot point, and will have the form of $mb^2$, where $b$ is a geometric size and mass distribution factor. Any rigid object you want to consider can have its moment of inertia put in that form. From this we see quickly that the actual mass of the object disappears: $$\frac{d^2\theta}{dt^2} = \frac{gr}{b^2}\sin\theta.$$

All that remains is to find $b$ which depends only on how the mass is distributed, not how much mass is present.

We also see that this is not simple harmonic motion. While the factor $g$ is not constant, it only introduces an anharmonic factor of $$1-\frac{r\theta^2}{R_E}-\frac{r^2\theta^4}{4R_E^2}$$. The $\sin\theta$ term introduces a larger anharmonicity because $$\sin\theta\simeq \theta-\frac{\theta^3}{6} = \theta\left(1-\frac{\theta^2}{6}\right).$$

So we see that 1) the mass doesn't matter, but the distribution of mass does, 2) the variation in height producing a variation in gravitational field only has a $(r/R_E)\theta^2$ affect, 3) the amplitude of the angle due to the $\sin\theta$ term becomes important when $\theta > 0.1$ radian.

Considering point 2), most pendula have $r<10 m$ and $R_E = 6.38\times 10^6$ m.

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Well the easy way is that the mass has a opposite affect when the bob goes up again on the other side. The deacceleration and the acceleration will equal out so the period will always be the same what ever mass you have.

       O
/I\
/ I \
/  I  \
0  0   0
A     C     B


Here you have a diagram to represent it. You drop the pendulum at B and it accelerates until it hits C then it will slow down. The mass will increase the deacceleration. So the acceleration and the deacceleration will equal out.

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NOOOOO!!! This is wrong. Granted the pendulum formula (T = 2 * pi * sqrt(L / g)) does not take into account mass of the bob, much less the pendulum, mass can and does affect the pendulum period. The Pendulum Formula is accurate and i give it credit, but its variables are broadly defined. T represents time or period, and g represents gravitational acceleration. I have no problem with those, but it's L that bothers me. To assume L is the distance from the point of axis to the bottom tip of the pendulum is to assume that the pendulum has an equal density throughout and its center of gravity lies directly in the center of the pendulum. However, with most pendulums this isn't the case. The bob, or weight on the pendulum, affects the location of the center of gravity. When a bob is added below the center of balance, the gravitational center of the entire pendulum is shifted downwards to some degree. Instead of saying L = the length of the pendulum, it's better to say that L = 2 * (distance between center of balance and pivot point).

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NOOOOO!!! This is wrong. – Jimmy360 Jun 14 at 21:56
This post is misleading. The formula for the period of the pendulum is correct, and is obtained assuming constant gravitational field. This is generally correct, but would be a problem for the case of very large pendulum, which clears the original question. Usually L is the distance to pendulum's center of mass and when a blob is involved, is usually assumed to contain the whole mass of the pendulum. But these are not considerations leading to the mass invariant formula for T. – rmhleo Jun 14 at 22:09
Certainly for real pendulums the ratio of the mass of the bob and the mass of the support affects the position of the center of mass of the device. But if you read closely any non-introductory account you'll see that L is defined the distance from the pivot to the CoM. An important detail, but not one that is appropriate to spend much time on in the first introduction (which usually assumes a massless rod). – dmckee Jun 14 at 23:52