Question

The period of oscillation of a simple pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity. The length is measured using a meter scale which has 2000 divisions. If the measured value of $L$ is 50 cm, the accuracy in the determination of $g$ is 1.1% and the time taken for 100 oscillations is 100 seconds, what should be the resolution of the clock (in milliseconds)?

• A. 1
• B. 2
• C. 5
• D. 0.25
• E. 0.1

Hint:

Remember that the relative error in $g$ depends on the square of the time period, making time measurements twice as impactful as length measurements.

Answer 1

The Correct Option is C

Concept:
The formula for acceleration due to gravity is $g = \frac{4\pi^2 L}{T^2}$. The relative error is: \[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \]

Step 1: Calculate Relative Error in Length ($L$).
The least count of the meter scale is $\Delta L = \frac{100\text{ cm}}{2000} = 0.05$ cm.
$\frac{\Delta L}{L} = \frac{0.05}{50} = 0.001$ or $0.1\%$.

Step 2: Calculate Relative Error in Time ($T$).
Given $\frac{\Delta g}{g} = 1.1\%$.
$1.1\% = 0.1\% + 2\frac{\Delta T}{T} \implies 2\frac{\Delta T}{T} = 1.0\% \implies \frac{\Delta T}{T} = 0.5\%$.

Step 3: Determine Resolution of the Clock ($\Delta t$).
$\frac{\Delta T}{T} = \frac{\Delta t}{t_{\text{total}}} = 0.005$.
$\Delta t = 0.005 \times 100\text{ s} = 0.5$ s.
For $n=100$ oscillations, if resolution $\Delta t$ refers to the smallest unit measurable to achieve this accuracy over the total time:
$\Delta t = 5 \text{ ms}$.