harmonic-oscillatornewtonian-gravitynewtonian-mechanicstime
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?
One of the reason's I like to cite for this is empirical observation. For years we have observed that the planets in the solar system move in closed orbits (roughly atleast). Mind you, I haven't asked for the orbits to be circular or elliptic, just closed. Now, invoking Bertrand's theorem with regard to the Kepler problem of the motion of a body in a central force, one can show that the only cases for which closed orbits are stable are that of the inverse square law and Hooke's law for the force. I find it kind of remarkable that the requirement of stability of closed orbits is sufficient to constrain the form of the central force to just two cases. Now the Hookean force is rather unphysical as it predicts increasingly large forces for large separations. Hence, by using this simple observation, we have sufficient reason to atleast believe that gravity acts (classically, that is) as an inverse square law force.
Suppose we take two identical pendulums. The green one is undamped and the red one is damped:
The force $F_g$ on the green pendulum bob is (approximately) given by the usual simple harmonic oscillator law:
$$ F_g = -kx $$
where $x$ is the displacement, and the acceleration is just $F_g/m$.
Now consider the red pendulum bob. This is damped, so the force on the bob will be the SHO force minus some damping force:
$$ F_r = -(kx - d)$$
where $d$ is some function of the bob velocity and possibly position. Again the acceleration of the red bob is $F_r/m$.
The point is that $F_g > F_r$ (more precisely the magnitude of $F_g$ is greater than the magnitude of $F_r$) and that means the acceleration of the green bob is greater than the acceleration of the red bob. So if we start the two bobs at the same point at time $t = 0$ the red bob will take longer to reach the centre because its acceleration is lower. But if the red bob takes longer to reach the centre than the green bob that must mean its period is longer i.e. its angular frequency is lower.
And that's why damping increases the period of the oscillation.
Best Answer
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.