How does the time period of a simple pendulum with a metallic bob vary if we heat its metallic bob? The pendulum is assumed to be a simple one and air drag is taken to be negligible.

Please provide a proof if possible.


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  • $\begingroup$ Welcome on Physics SE :) What do you expect to happen? Did you have any ideas about this yourself? $\endgroup$ – Sanya Aug 6 '16 at 7:28
  • $\begingroup$ That depends on how the bob is connected to the fiber. If the fiber is holding the bob at top, the effective length will increase, if it's holding it at the bottom, it will decrease. If it's done intelligently, the temperature coefficients of fiber and bob could even compensate each other, at least in first order. In practice the fiber will have the greater problems, though, which leads to the following design: en.wikipedia.org/wiki/Gridiron_pendulum $\endgroup$ – CuriousOne Aug 6 '16 at 7:28
  • $\begingroup$ Yes... but the difference is dwarfed by approximations made in idealizing the pendulum problem, under most circumstances. $\endgroup$ – anon01 Aug 6 '16 at 7:37
  • $\begingroup$ if the CM of the bob shifts w.r.t the suspension point , frequency will change. You can do the rest yourself. $\endgroup$ – Lelouch Aug 6 '16 at 9:09
  • $\begingroup$ Can you please elaborate? $\endgroup$ – shivam mishra Aug 6 '16 at 9:11

Let's assume the length of the string remains constant, i.e. only the bob is affected by the heating. This appears to what the question implies. In that case to a first approximation there will be no effect on the period. The period is given by:

$$ \tau = 2\pi\sqrt{\frac{\ell}{g}} $$

and since neither $\ell$ nor $g$ are affected by heating the period does not change.

However I can think of two ways in which the period could change very slightly due to heating, and both are related to the thermal expansion of the bob.

Effect on the bouyant force

Firstly we assume that the downwards force on the bob is given by:

$$ F_d = mg $$

However there is a small bouyant force due to the volume of air displaced by the bob, so the force is actually:

$$ F_d = mg - V\rho g $$

and taking this into account the period is:

$$ \tau = 2\pi\sqrt{\frac{m\ell}{(m-\rho V)g}} $$

where $\rho$ is the density of the air and $V$ is the volume of the bob. If we heat the bob its volume will increase due to thermal expansion so the period will increase.

Effect on the moment of inertia

The other affect arises from the approximation that we treat the bob as a point mass. The equation of motion of the pendulum is:

$$ T = I\ddot{\theta} $$

where $T$ is the torque, $I$ is the moment of inertia and $\theta$ is the angle of rotation. For a point mass, and using the small angle approximation, $T=mg\ell\theta$ and $I=m\ell^2$ so we get:

$$ mg\ell\theta = m\ell^2\ddot{\theta} $$

which rearranges to our usual equation:

$$ \ddot{\theta} = \frac{g}{\ell}\theta $$

However suppose the bob is a sphere or radius $r$. The moment of inertia of a sphere about an axis though the centre is:

$$ I_\text{sphere} = \tfrac{2}{5}mr^2 $$

So using the parallel axis theorem the moment of inertia about the axis a distance $\ell$ away is:

$$ I = \tfrac{2}{5}mr^2 + m\ell^2 $$

In that case the equation of motion is:

$$ mg\ell\theta = -m\left( \tfrac{2}{5}r^2 + \ell^2 \right)\ddot{\theta} $$ and rearranging gives:

$$ \ddot{\theta} = -\frac{g\ell}{\tfrac{2}{5}r^2 + \ell^2}\theta $$

so our period becomes:

$$ \tau = \sqrt{\frac{\tfrac{2}{5}r^2 + \ell^2}{g\ell}} $$

If we heat the bob its radius $r$ will increase due to the thermal expansion so the period will increase with heating.


So both the effects discussed above mean the period will increase if the bob expands due to heating. However while this sort of exercise is fun to do I must emphasise that these effects will be tiny and in most cases dwarfed by other approximations such as the small angle approximation and ignoring air resistance.


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