A student carries out an experiment to find the time period of a simple pendulum. The student records three times for the pendulum to complete 10 oscillations:
\begin{tabular}{ccc}
14.75 s & 13.05 s & 13.50 s \\
[tex]$t_1$[/tex] & [tex]$t_2$[/tex] & [tex]$t_3$[/tex]
\end{tabular}

(i) Calculate the average time for 10 oscillations.
(ii) Determine the time period.
(iii) Calculate the frequency.



Answer :

Let's solve the problem step-by-step.

### (i) Average time for 10 oscillations
The student recorded the times for 10 complete oscillations three times: [tex]\( t_1 = 14.75 \)[/tex] seconds, [tex]\( t_2 = 13.05 \)[/tex] seconds, and [tex]\( t_3 = 13.5 \)[/tex] seconds.

To find the average time for 10 oscillations, we sum these three times and divide by 3:
[tex]\[ \text{Average time for 10 oscillations} = \frac{t_1 + t_2 + t_3}{3} = \frac{14.75 + 13.05 + 13.5}{3} = 13.766666666666666 \text{ seconds} \][/tex]

### (ii) Time period
The time period [tex]\( T \)[/tex] is defined as the time it takes for one complete oscillation. To find the time period from the average time for 10 oscillations, we divide the average time by 10:
[tex]\[ \text{Time period} = \frac{\text{Average time for 10 oscillations}}{10} = \frac{13.766666666666666}{10} = 1.3766666666666665 \text{ seconds} \][/tex]

### (iii) Frequency
The frequency [tex]\( f \)[/tex] is defined as the number of oscillations per unit time and is the reciprocal of the time period:
[tex]\[ \text{Frequency} = \frac{1}{\text{Time period}} = \frac{1}{1.3766666666666665} = 0.7263922518159808 \text{ Hz} \][/tex]

To summarize:
1. The average time for 10 oscillations is [tex]\( 13.766666666666666 \)[/tex] seconds.
2. The time period for one oscillation is [tex]\( 1.3766666666666665 \)[/tex] seconds.
3. The frequency of the pendulum's oscillations is [tex]\( 0.7263922518159808 \)[/tex] Hertz.