Question

How do I calculate the standard error of g in T=2π√l/g

Answer 1

Answer:

T=2πlg−−√

or T=(2π)(l)+1/2(g)−1/2

Taking logarithm of both sides, we have

In (T)=ln(2π)+12(Inl)−(12)ln(g) ………….(i)

Here, 2π is a constant, therefore In (2π) is also a constant.

Differentiating Eq. (i), we have

1TdT=0+12(ll)(dl)−12(12)(dg)

or (dtT)max= Maximum value of (±12dll−+12dgg)

=12(dll)+12(dgg)

This can also be written as

or(ΔTT×100)max=12[(Δll)×100]+12[Δgg×100] or percentage error in time period

=±[12(percentage error inl)+12(percentage error ing)]

=±[12×1+12×2]=±1.5%.

Explanation: