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

Please correct me if I'm going wrong -
By the gravitation formula: $F = \frac{G m_1 m_2}{r^2}$, So if the mass of a bob is greater then the torque on it should increase because the Force increased now by a very small magnitude. So it should go faster and thus the Time period should be lesser.

But my Physics book says that Time period is only affected by effective length and $g$ Why doesn't mass of bob affect it?

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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

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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