Question

Calculate the maximum percentage error in the measurement of time for a pendulum. The time taken by the pendulum to complete $20$ oscillations is $25$ seconds. Also, the least count of the stopwatch is? second.

Answer 1

Step 1: Given data

Number of oscillations $=20$

Time $\left(\mathrm{t}\right)=25$

Step 2: Formula used

Actual $=$ Observed $-\mathrm{LC}$

$\mathrm{LC}=$Least count

Step 3: Determine the actual time

The minimal remember of a tool is described because the minimal, and the minimal correct cost that may be measured with the aid of using a sure tool is called the tool`s minimal remember.

The minimal remember for each tool isn't always steady and could range due to production faults and different issues that arise at some point of processing.

In the question given, it is designated as a stopwatch with a minimum value of $0.2$. this is, the smallest and most accurate value that can be measured with this stopwatch is $0.2$.

In a given experiment, Bob has been observed to vibrate $20$ times, taking $25$ seconds to complete the $20$ vibrations.

Here, the actual value it takes for Bob to swing $20$ times is shown as the difference between the time obtained and the minimum count of the device.

Actual $=$ observed $-\mathrm{LC}$

$\mathrm{Actual}=25-0.2\mathrm{s}=24.8\mathrm{s}$

Here, the time error is calculated as the ratio of change over time to the observed value.

Step 4: Determine the maximum percentage

This means that it is the ratio of the instrument's minimum count to the observed value.

$\frac{\mathrm{\Delta T}}{\mathrm{T}}=\pm \left(\frac{\mathrm{LC}}{\text{Observed}}\right)$

Substituting the values ​​and converting them to percentiles looks like this:

$\left(\frac{\mathrm{\Delta T}}{\mathrm{T}}\times 100\right)=\pm \left(\frac{0.2}{25}\times 100\right)=\pm 0.8\%$

The maximum percentage error in the measurement of time for a pendulum is $\pm 0.8\%$..