Question

A student uses a stopwatch to measure the period of the pendulum of the Beverly clock in the corridor. His measurements are: (a) 13.54is (b) 13.44 s (c) 13.89 s (d) 13.41 s (e) 13.17s (f) 13.22 s . Calculate his best estimate of the period Calculate the associated error .Explain the assumptions behind your estimation of the error

Answer 1

Answer:

Reading is close to (b) 13.44 which is the best estimate of the period

Associated error, $\Delta E =0.178 s$

Given:

$t_{a} = 13.54 s$

$t_{b} = 13.44 s$

$t_{c} = 13.89 s$

$t_{d} = 13.41 s$

$t_{e} = 13.17 s$

$t_{f} = 13.22 s$

Solution:

1.The best estimate of the period can be calculated by the mean of the measurements and the one closest to the mean is the best estimate of the measurement:

$Mean, \bar {x} = \fra{sum of all observations}{No. of observation}$

$Mean, \bar {x} = \frac{t_{a} + t_{b} + t_{c} +t_{d} + t_{e} + t_{f}}{6}$

$Mean, \bar {x} = \frac{13.54 + 13.44 + 13.89 + 13.41 + 13.17 + 13.22}{6}$

$Mean, \bar {x} = 13.445 s$

It is close to 13.44 s

2. Associated error is given by:

$\Delta E_{n} = |measured value - actual value|$

$\Delta E_{n} = |t_{n} - \bar {x}|$

where

n = a, b,......, e

Now,

$\Delta E_{a} = |t_{a} - \bar {x}| = |13.54 - 13.44| = 0.01$

$\Delta E_{b} = |t_{b} - \bar {x}| = |13.44 - 13.44| = 0.00$

$\Delta E_{c} = |t_{c} - \bar {x}| = |13.89 - 13.44| = 0.45$

$\Delta E_{d} = |t_{d} - \bar {x}| = |13.41 - 13.44| = 0.03$

$\Delta E_{e} = |t_{e} - \bar {x}| = |13.17 - 13.44| = 0.027$

$\Delta E_{f} = |t_{f} - \bar {x}| = |13.54 - 13.44| = 0.10$

Mean Absolute Error, $\Delta E = \frac{\Sigma E_{n}}{6}$

$\Delta E = \frac{0.01 + 0.00 + 0.45 + 0.03 +0.027 + 1.10}{6}$

$\Delta E =0.178 s$

3. The assumption behind the estimation is population is considered to distributed normally.