The oscillation frequency of a pendulum in the small angle limit: $$\omega = \sqrt\frac{l}{g} $$
Now I am trying to work out the new frequency if the pendulum is immersed in a zero viscosity fluid and I have run into a conceptual problem.
For a general sphere radius a moving in a fluid we have an added mass of $ m_a = \frac{2\pi a^3 \rho_0}{3} $ where $\rho_0$ is the density of the fluid. This added mass is due to the motion of the fluid the sphere drags with it.
Moreover, there is a buoyancy force decreasing the effective mass with $ \frac{4\pi a^3 \rho_0}{3} $ so the overall effective mass is $$m_{effective} = m +\frac{2\pi a^3 \rho_0}{3}- \frac{4\pi a^3 \rho_0}{3} = m – \frac{2\pi a^3 \rho_0}{3} = \frac{2\pi a^3 (2\rho-\rho_0)}{3}$$
Now if we would be looking for the new frequency of a mass on a spring we would simply plug this into $\omega = \sqrt(k/m)$ and have an answer.
However, the pendulum is different in that the mass takes no role in the basic equation. Intuitively, I still think the period should change, so I have tried deriving the pendulum equation with the extra effects included, but I am not sure how to include the added mass due to the motion of the fluid:
Should $m_a$ be included in the weight of the the pendulum? Perhaps not because it is supported by bouyancy?
But should it be included in the moment of inertia?
Best Answer
The buoyancy modifies the effective gravity experienced by the pendulum: $g'=g(\rho-\rho_0)/\rho$. The equation of motion would be: ($V$ is bob volume) $$(\rho+\rho_0/2)Vl^2\ddot{\theta}=-\rho Vg'l\sin\theta\\$$
Then for small amplitude motion, the period: $$\tau=\sqrt{\frac{l}{g}\left( \frac{\rho+\rho_0/2}{|\rho-\rho_0|}\right)}$$