If the Length of a Clock Pendulum Increases by 0.2%, What Happens?

The time period (T) of a simple pendulum is given by the formula: $T = 2\pi \sqrt{\frac{L}{g}}$

• T = Time taken for one complete swing.
• L = Length of the pendulum string/rod.
• g = Gravity (which is a constant).
• Notice that Time ($T$) is directly proportional to the square root of Length ($\sqrt{L}$). Therefore, if the length increases, the time period MUST increase.

To find the small percentage change, we use the fractional error formula (taking logarithms and differentiating): $\frac{\Delta T}{T} = \frac{1}{2} \times \frac{\Delta L}{L}$

The question states the length increased by 0.2% ($\frac{\Delta L}{L} = 0.2%$). Let's plug it into the formula:

• Percentage change in Time = $\frac{1}{2} \times 0.2%$
• Percentage change in Time = 0.1%

• Result: The time period of the pendulum increases by exactly 0.1%.
• What does this physically mean?: Because the length is longer, the pendulum takes more time to finish one swing. Because every swing is suddenly slower, the clock will start running 'Slow' and will ultimately lose time over the course of the day. You will be late for your appointments!