Animated Solution for Physics - Oscillations: The bob of a simple pendulum executes simple harmonic motion in water with a period $t$, while the period of oscillation of the bob is $t_0$ in air. Neglecting frictional force of water and given that the density of the bob is $\frac{4}{3}\times 1000{\text{ kgm}}^{-3}$. What relationship between $t$ and $t_0$ is true?

Pendulum in Air

Imagine a simple pendulum oscillating freely in the air. In this ideal scenario, the only downward force acting on the bob is its own weight due to gravity, $mg$.

The tension in the string provides the necessary restoring force, giving us the classic time period formula:

This is our baseline. The pendulum swings with a steady, predictable rhythm dictated purely by the length of the string and the acceleration due to gravity.

Immersion in Water

Now, let's make things interesting. We fill the container with water, completely immersing the pendulum.

The water isn't just a passive background; it actively interacts with the bob. According to Archimedes' principle, the water exerts an upward buoyant force ($F_B$) on the pendulum's bob, directly opposing gravity.

Because of this upward buoyant force, the bob feels lighter. Its effective weight decreases. We can capture this physical reality by defining a new effective acceleration due to gravity, $g_{\text{eff}}$.

The net downward force on the bob is now $mg-F_B$.

Dividing this by the mass $m$ gives us the effective gravity:

We know that the buoyant force is the weight of the displaced fluid, $F_B={\rho }^{\prime }Vg$, where ${\rho }^{\prime }$ is the density of the fluid and $V$ is the volume of the bob. The mass of the bob is $m=\rho V$. Substituting these in, we get:

This elegant equation tells us that the effective gravity depends entirely on the ratio of the fluid's density to the bob's density.

To find this effective gravity, we need to substitute the given density values.

The density of water, ${\rho }^{\prime }$, is a standard $1000{\text{ kg/m}}^3$. The density of the bob, $\rho$, is given as $\frac{4}{3}\times 1000{\text{ kg/m}}^3$. Let's calculate their ratio:

Now, let's bring that ratio back into our $g_{\text{eff}}$ equation:

This is a massive reduction! The effective gravity acting on the bob in water is just one-fourth of the normal gravity.

With our new effective gravity calculated, let's write down the formula for the time period $t$ in water.

It has the exact same structure as before, but we must carefully substitute $\frac{g}{4}$ in place of the standard $g$:

Look at the denominator inside the square root. That $4$ will come out of the square root as a $2$ in the numerator:

And what is $2\pi \sqrt{\frac{l}{g}}$? That is exactly our original time period, $t_0$!

So, the final answer is:

$$t=2t_0$$

The pendulum swings twice as slow in the water.

Before we wrap up, think about this: what would happen if the liquid's density was actually greater than the bob's density? The buoyant force would overpower gravity, the bob would float to the surface, the string would go slack, and simple harmonic motion wouldn't even be possible! Always keep these physical constraints in mind.