Question

We measure the period of
oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71s.and 2.80s. Calculate the absolute errors,relative error or percentage error.​

Answer 1

Answer:

We measure the period of

oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71s.and 2.80s

Answer 2

To Find :

• absolute errors , relative error or percentage error.

Solution :

Reading : 2.63 s , 2.56 s, 2.42 s, 2.71 s and 2.80 s.

The mean period of oscillation pendulum

$\sf \:T \: = \frac{sum \: of \: all \: observations}{number \: of \: observations} \\ \\ \sf \:T \: = \frac{(2.63 + 2.56 + 2.42 + 2.71 + 2.80)s}{5} \\ \\ \sf \:T \: = \frac{13.12 \: s}{5} \\ \\ \sf \:T \: = 2.624 \: s \\ \\ \sf \:T \: = 2.62 \: s$

• Errors In measurements are :-

2.63 s - 2.62 s = 0.01 s

2.56 s - 2.62 s = - 0.06 s

2.42 s - 2.62 s = 0. 20 s

2.71 s - 2.62 s = 0.09 s

2.80 s - 2.62 s = 0.18 s

Arithmetic mean of all absolute errors :

$\Delta\sf \:T \: = \frac{(0.01 + 0.06 + 0.20 + 0.09 + 0.18)s}{5} \\ \\ \sf \: = \frac{0.54 \: s}{5} \\ \\ \sf = 0.11 \: s$

T = 2.6 ± 0.1 s

The relative error or percentage error :

$\sf\:\Delta{a} = \frac{0.1}{2.6}\times 100\\\\\sf = 4\%$

━━━━━━━━━━━━━━━━━━━━━━━━━