Understanding the Problem

We are given a simple pendulum of length $l=1\text{m}$ used to determine $g$, the acceleration due to gravity. The time for 20 oscillations is measured as 40 seconds using a stopwatch with least count 1 second. We need to analyze the errors in this measurement and determine which statements are true.

Step 1: Recall the Formula for Time Period

The time period $T$ of a simple pendulum is given by:

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

Squaring both sides and rearranging for $g$:

$g=\frac{4{\pi }^{2}l}{T^2}$

Step 2: Calculate the Time Period $T$

Time for 20 oscillations = 40 s

So, time period $T=\frac{40}{20}=2\text{s}$

Step 3: Determine the Error in Measuring Time Period ($\mathrm{\Delta T}$)

The stopwatch has a least count of 1 second. This error is for the total time measured (40 s for 20 oscillations).

Error in total time $\mathrm{\Delta t}=1\text{s}$

Since $T=\frac{t}{n}$, where $t=40\text{s}$ and $n=20$,

The error in $T$ is $\mathrm{\Delta T}=\frac{\mathrm{\Delta t}}{n}=\frac{1}{20}=0.05\text{s}$

So, $\mathrm{\Delta T}=0.05\text{s}$ is correct.

Step 4: Calculate Percentage Error in $g$

From $g=\frac{4{\pi }^{2}l}{T^2}$, we find the relative error:

$\frac{\mathrm{\Delta g}}{g}=\frac{\mathrm{\Delta l}}{l}+2\frac{\mathrm{\Delta T}}{T}$

Given $l=1\text{m}$ (exact, so $\mathrm{\Delta l}=0$) and $T=2\text{s}$, $\mathrm{\Delta T}=0.05\text{s}$

So, $\frac{\mathrm{\Delta g}}{g}=0+2\times \frac{0.05}{2}=2\times 0.025=0.05$

Percentage error in $g$ = $0.05\times 100=5\%$

So, percentage error in g is 5% is correct.

Final Answer

The true statements are:

• Percentage error in the determination of g is 5%
• Error ΔT in measuring T, the time period, is 0.05 seconds

Related Topics

Simple Pendulum: A mass attached to a string that oscillates under gravity. Used to measure g.

Error Analysis: Study of uncertainties in measurements. Important in experiments.

Least Count: The smallest value that can be measured by an instrument.

Important Formulae

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

Gravity: $g=\frac{4{\pi }^{2}l}{T^2}$

Relative error in g: $\frac{\mathrm{\Delta g}}{g}=\frac{\mathrm{\Delta l}}{l}+2\frac{\mathrm{\Delta T}}{T}$