Question

in an experiment of simple pendulum, IIT student made several observations for the period of oscillation. He is reading turned out to the period of oscillation.His reading turned out to be: 2.63sec, 2.56sec, 2.42sec, 2.71sec, and 2.80sec. with the help of above observations calculate absolute errors and relative error.​

Answer 1

First, we derive the Artithmetic mean of the given observations.

$A_m_e_a_n$ = $\frac{sum-of-all-observations}{number-of-observations}$

= $\frac{2.63+2.56 + 2.42+2.71+2.80}{5} secs$

$=\frac{13.12}{5} secs = 2.62 secs$

Calculating Absolute Error :-

delta $A_1$ = $(2.63 - 2.62) secs$ = $0.01 secs$

delta $A_2$ = $(2.56 - 2.62) secs$ = $-0.06 secs$

delta $A_3$ = $(2.42 - 2.62) secs$ = $-0.20 secs$

delta $A_4$ = $(2.71 - 2.62) secs$ = $0.09 secs$

delta $A_5$ = $(2.80 - 2.62) secs$ = $0.18 secs$

For the following calculation, we will not consider the presence of the minus (-) in cases of $A_2$ and $A_3$

∴ delta $A_m_e_a_n =$ (|delta $A_1$| + |delta $A_2$| + |delta $A_3$| + |delta $A_4$| + |delta $A_5$|) ÷ 5 secs = [0.108 ] secs

= 0.11 secs

Calculating Relative Error :-

We use the following formula;

$\frac{Mean-Absolute-Error}{A_m_e_a_n}$  $= \frac{0.11}{0.26} secs$ = 0.04 secs

Ans) Absolute Error = 0.11 secs; Relative error = 0.04 secs

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Answer 2

Given data :

When a student made observations for the period of oscillations of a simple pendulum, the corresponding readings noted are :

2.63 sec, 2.56 sec, 2.42 sec, 2.71 sec and 2.80 sec.

Calculating Absolute mean:

Before jumping into the calculation of absolute error, we need to find out the Absolute mean.

$AbsoluteMean=\frac{Sum of Observations}{no.of observations}$

$Abs_{Mean} =\frac{reading_{1}+reading_{2}+...+reading_{n} }{n}$

              $= \frac{(2.63 + 2.56 + 2.42 + 2.71 + 2.80)}{5}$

              $= \frac{13.12}{5}$

$Abs_{mean}$ = 2.62 secs

So, the mean period of oscillating time is 2.62 secs.

Calculating Absolute Errors :

Now, we need to calculate the absolute errors of means. All you need to do is; to find the absolute differences i.e., modulus differences between them. As in,

ΔA₁ = I 2.63 - 2.62 I = 0.01 secs

ΔA₂ = I 2.56 - 2.62 I = 0.06 secs

ΔA₃ = I 2.42 - 2.62 I = 0.2 secs

ΔA₄ = I 2.71 - 2.62 I = 0.09 secs

ΔA₅ = I 2.80 - 2.62 I =0.18 secs

Absolute Mean Error = (ΔA₁ + ΔA₂ + ΔA₃ + ... + ΔAₙ) / n

                                   = (ΔA₁ + ΔA₂ + ΔA₃ + ΔA₄ + ΔA₅) / 5

                                   = (0.01 + 0.06 + 0.2 + 0.09 + 0.18 ) / 5

                                   = 0.54 / 5

                                   = 0.108

                                   = 0.11 secs

Therefore, Absolute mean error is 0.11 secs

Calculation of time period per oscillation :

As the absolute mean error period of the oscillation is 0.11 sec, the timpe period for an oscillation of a simple pendulum lies between (Absolute mean - Absolute mean error) and (Absolute mean + Absolute mean error).

That is, the time period of the simple pendulum's oscillation is in between 2.62 ± 0.11 secs.

Calculation of Relative error :

$RelativeError = \frac{Absolute Error}{Absolute Mean}$

                        $= \frac{0.11}{0.24}$

                         = 0.04 secs (rounded)

As Relative Error is 0.04 secs,

Relative Error Percentage = (0.04 x 100)% = 4%

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