- Home
- Standard 11
- Physics
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be $2.63 \;s , 2.56 \;s , 2.42\; s , 2.71 \;s$ and $2.80 \;s$. Calculate the absolute errors, relative error or percentage error.
Solution
Answer The mean perlod of oscillation of the pendulum
$T \;=\frac{(2.63+2.56+2.42+2.71+2.80) \,s}{5}$
$\quad=\frac{13.12}{5} \;s$
$=2.624\, s $
$=2.62 \,s$
As the periods are measured to a resolution of $0.01 \,s ,$ all times are to the second decimal; it is proper to put this mean perlod also to the second decimal.
The errors in the measurements are
$2.63 \,s -2.62 \,s =0.01 \,s$
$2.56 \,s-2.62 \,s=-0.06 \,s$
$2.42\, s -2.62\, s =-0.20 \,s$
$2.71 \,s -2.62\, s =0.09 \,s$
$2.80\, s-2.62\, s=0.18\, s$
The arthmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) is
$ \Delta T_{\text {mean}} =[(0.01+0.06+0.20+0.09+0.18) \,s ] / 5 $
$=0.54 \,s / 5 $
$=0.11 \,s $
$T=2.6 \pm 0.1 \,s$
$\delta a=\frac{0.1}{2.6} \times 100=4 \%$
