A bob of mass m suspended by a thread of length l undergoes simple harmonic oscillations with time p

Q: Why does the gravity change when the pendulum is in a liquid?

Gravity itself doesn't change, but the *net downward acceleration* of the bob changes. The liquid exerts an upward buoyant force (Archimedes' principle) that opposes gravity. This makes the bob feel 'lighter', which we mathematically represent as a reduced 'effective gravity'.

Q: How did we get the formula $g_{\text{eff}} = g(1 - \rho/\sigma)$?

The net downward force is Weight - Buoyant Force. So, $mg_{\text{eff}} = mg - V\rho g$. Since mass $m = V\sigma$, we can write $V\sigma g_{\text{eff}} = V\sigma g - V\rho g$. Dividing the entire equation by $V\sigma$ gives $g_{\text{eff}} = g - g(\rho/\sigma) = g(1 - \rho/\sigma)$.

Q: What would happen if the liquid was denser than the bob?

If the liquid's density ($\rho$) was greater than the bob's density ($\sigma$), the buoyant force would be stronger than gravity. The bob would float to the surface! It wouldn't oscillate as a simple pendulum hanging downwards anymore.

Q: I always mess up the fraction division inside the square root. Any tips?

Write it out explicitly as a division problem: $(4l/3) \div (3g/4)$. Then, remember the rule 'keep, change, flip'. Keep the first fraction $(4l/3)$, change division to multiplication, and flip the second fraction to $(4/3g)$. Multiplying them gives $(4l \cdot 4) / (3 \cdot 3g) = 16l / 9g$.

Q: Does the mass of the bob affect the time period?

No! Notice that the mass $m$ does not appear in the final formula $T = 2\pi\sqrt{l/g}$. As long as the density ratio remains the same, a heavier or lighter bob of the same material will have the exact same time period.

Question

A bob of mass $m$ suspended by a thread of length $l$ undergoes simple harmonic oscillations with time period $T$. If the bob is immersed in a liquid that has density $1/4$ times that of the bob and the length of the thread is increased by $1/3$rd of the original length, then the time period of the simple harmonic oscillations will be

Options:

(A) $T$

(B) $\frac{3}{2}T$

(C) $\frac{3}{4}T$

(D) $\frac{4}{3}T$

Sigma Percentile · Accepted Answer

### The Physics of a Pendulum in a Fluid

Imagine a simple pendulum swinging freely in the air. Its motion is governed by gravity pulling it down and the tension in the string keeping it in a circular arc. The time period of this oscillation is a classic result in physics:

$$T = 2\pi \sqrt{\frac{l}{g}}$$

But what happens when we take this entire setup and submerge it in a liquid? The environment changes drastically. The liquid isn't just empty space; it exerts an upward force on the bob known as the **buoyant force**.

### The Concept of Effective Gravity

Because the buoyant force pushes upwards, it directly opposes the downward pull of gravity. From the perspective of the pendulum bob, it feels as though gravity itself has become weaker. We call this reduced gravitational pull the **effective gravity** ($g_{	ext{eff}}$).

By analyzing the forces (Weight downwards, Buoyancy upwards), we can derive a beautiful relationship for effective gravity:

$$g_{	ext{eff}} = g\left(1 - \frac{ho_{	ext{liquid}}}{\sigma_{	ext{bob}}}ight)$$

In our specific problem, we are given a crucial piece of information: the density of the liquid is exactly one-fourth the density of the bob.

$$ho_{	ext{liquid}} = \frac{1}{4}\sigma_{	ext{bob}}$$

Substituting this into our effective gravity equation, we find:

$$g_{	ext{eff}} = g\left(1 - \frac{1}{4}ight) = \frac{3g}{4}$$

So, the pendulum behaves as if it's on a planet where gravity is only $75\%$ as strong as Earth's!

### Adjusting the Length

The problem adds another twist: the length of the thread is increased by one-third of its original length. Let's calculate the new length, $l'$:

$$l' = l + \frac{l}{3} = \frac{4l}{3}$$

### The Grand Synthesis

Now we have all the pieces of the puzzle. We have a new length ($l'$) and a new effective gravity ($g_{	ext{eff}}$). Let's plug these into our master time period equation to find the new time period, $T'$:

$$T' = 2\pi \sqrt{\frac{l'}{g_{	ext{eff}}}}$$

Substituting our calculated values:

$$T' = 2\pi \sqrt{\frac{4l/3}{3g/4}}$$

Here is where we must be careful with our algebra. When dividing fractions, we multiply by the reciprocal. The $4$ in the denominator's denominator flips up to multiply the numerator's $4$, giving $16$. The $3$ in the numerator's denominator multiplies the denominator's $3$, giving $9$.

$$T' = 2\pi \sqrt{\frac{16l}{9g}}$$

We can pull the fraction $\frac{16}{9}$ out of the square root as $\frac{4}{3}$:

$$T' = \frac{4}{3} \left(2\pi \sqrt{\frac{l}{g}}ight)$$

Look closely at the term in the parentheses. It is exactly our original time period, $T$! Therefore, we arrive at our final, elegant conclusion:

$$T' = \frac{4}{3}T$$

The time period has increased, which makes perfect physical sense: a longer string and weaker effective gravity both contribute to a slower, more leisurely swing.

Select Answer:

Visualized Solution

Initial Time Period

Effective Gravity in Liquid

Calculating $g_{eff}$

New Length of Pendulum

Substituting into New Time Period

Simplifying the Expression

Final Answer

The Sigma Insight: Simple Harmonic Motion (SHM)

The Physics of a Pendulum in a Fluid

The Concept of Effective Gravity

Adjusting the Length

The Grand Synthesis

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