Question

A person measures the time period of 50 osciations of a simple pendulum 5 times
The data set is 46 s. 44.5 ,47.5 5.45 s and 47 s. It least count of the measuring
clock is 0.5 s, then the reported ume period should be​

Answer 1

Given:

A person measures the time period of 50 osciations of a simple pendulum 5 times  

To find:

The reported time period should be​

Solution:

From given, we have,

A person measures the time period of 50 osciations of a simple pendulum 5 times .

The data set is 46 s, 44.5 ,47.5 54.5 s and 47 s.

Mean value of the time = (46 + 44.5 + 47.5 + 54.5 + 47)/5 = 47.9 s

tm = 47.9 s

The absolute error is calculated as follows.

Δt = |tm - t1| = |47.9 - 46| = 1.9 s

Δt = |tm - t1| = |47.9 - 44.5| = 3.4 s

Δt = |tm - t1| = |47.9 - 47.5| = 0.4 s

Δt = |tm - t1| = |47.9 - 54.5| = 6.6 s

Δt = |tm - t1| = |47.9 - 47| = 0.9 s

The mean absolute error is calculated as follows.

Δt = (1.9 + 3.4 + 0.4 + 6.6 + 0.9) / 5 = 2.64 s

Δt ≈ 2.6 s

The mean time is (47.9 ± 2.6) s