Animated Solution for Physics - Oscillations: A pendulum clock loses 12 s a day if the temperature is $40^{\circ }\text{C}$ and gains 4 s in a day if the temperature is $20^{\circ }\text{C}$. The temperature at which the clock will show correct time, and the coefficient of linear expansion ($\alpha$) of the metal of the pendulum shaft are, respectively

Imagine a classic grandfather clock standing tall in a hallway. Its heartbeat, the rhythmic ticking, depends entirely on the length of its pendulum. The time period of a simple pendulum is given by the famous equation $T=2\pi \sqrt{\frac{L}{g}}$.

But here is the catch: the physical world is not static. Metals expand when heated and contract when cooled. So, on a sweltering hot day of $40^{\circ }\text{C}$, the pendulum rod lengthens. A longer pendulum means a larger time period $T$; the clock swings slower and consequently loses time. Conversely, on a cooler day of $20^{\circ }\text{C}$, the rod shrinks, the time period decreases, and the clock runs fast, gaining time.

To quantify exactly how much time is lost or gained, we look at the fractional change in the time period. Since $T\propto L^{1/2}$, a small fractional change in length results in half that fractional change in the time period:

From the physics of thermal expansion, we know that $\frac{\Delta L}{L}=\alpha \Delta \theta$, where $\alpha$ is the coefficient of linear expansion and $\Delta \theta$ is the change in temperature. Substituting this in, we get the master equation for the time error $\Delta t$ accumulated over a total time $t$:

Here, ${\theta }_0$ represents the "sweet spot"—the exact temperature at which the clock is perfectly calibrated and shows the correct time.

Let's assume the clock shows the exact correct time at an unknown temperature $\theta$. We are given two distinct scenarios.

Scenario 1: At $40^{\circ }\text{C}$, the clock loses 12 seconds in a day. This means $40^{\circ }\text{C}$ is hotter than our ideal temperature $\theta$. We can write:

Scenario 2: At $20^{\circ }\text{C}$, the clock gains 4 seconds in a day. This means $20^{\circ }\text{C}$ is colder than our ideal temperature $\theta$. We can write:

Notice how we carefully arranged the temperature differences to ensure they are positive magnitudes.

We now have a beautiful system of two equations. To eliminate the unknowns $\alpha$ and $t$, we simply divide equation (i) by equation (ii):

This simplifies elegantly to:

Cross-multiplying gives us a straightforward linear equation:

This is the sweet spot! At exactly $25^{\circ }\text{C}$, the clock will keep perfect time.

Now for the final piece of the puzzle: finding $\alpha$. Let's substitute our newly found $\theta =25^{\circ }\text{C}$ back into equation (ii):

Here is where many students make a silly mistake. The time error (4 seconds) is in seconds, so the total time $t$ (which is 1 day) must also be converted into seconds to maintain dimensional consistency.

Substituting this in:

This perfectly matches option (a).

Before we wrap up, think about this: how did ancient clockmakers solve this annoying temperature problem? They couldn't control the weather!

They invented the Gridiron pendulum. By using alternating rods of two different metals, like steel and brass, they cleverly designed the pendulum so that the upward expansion of one metal exactly canceled out the downward expansion of the other. The effective length remained perfectly constant, and the clock never lost a second. Physics is truly beautiful!