Why doesn't mass of bob affect time period?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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).