Question

The acceleration of gravity can be found from the length land period T of a pendulum; the formula is$\mathbf{g}\mathbf{=}\frac{\mathbf{4}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{l}}{{\mathbf{T}}^{\mathbf{2}}}$. Find the relative error ingin the worst case if the relative error inIis 5%, and the relative error inTis 2%.

Short answer

The answer is$g_{gre}=0.09$ ..

Step-by-step solution

Explanation of Solution

Provided:

The acceleration of gravity is$g=\frac{4{\pi }^{2}l}{T^2}$ ,Where I is length and T is pendulum.

Approximation by differentials

This method is based on the derivatives of functions whose values must be calculated at certain locations, as the name implies.

Consider a function $\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, find the value of a function $\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$when$\mathbf{x}\mathbf{=}\mathbf{x}\mathbf{\text{'}}$

As an example, the derivative of a function $\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$with respect to xwill be employed.

$\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{=}\mathbf{\left(}\mathbf{f}\mathbf{\right(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}$is Change in with respect to change in xas$\mathbf{dx}\mathbf{\to }\mathbf{0}\mathbf{.}$.

If the value of $x=x\text{'}$from a value of x near it, such that the difference in the two values, dx, is vanishingly small, one can derive the change in the value of the function $y=f\left(x\right)$ corresponding to the change dx in x . In practice, however, the concept of vanishingly small is not possible.

Calculation

Consider the relative errors for two of the three variables, differentials may be used to calculate the relative error in$g\left(g_{re}\right)$.

It will, however, need to first extract the relationship's differentials:

Theis g represented by the expression:

$g=\frac{4{\pi }^{2}l}{T^2}$

Take log on both sides,

$Ing=In\left(4{\pi }^2\right)+Inl-2InT$

Here,

$\frac{\mathrm{dg}}{\mathrm{g}}=\frac{\mathrm{dl}}{\mathrm{I}}-2\frac{\mathrm{dT}}{\mathrm{T}}$

In the worst case,$\frac{dT}{T}<0$.

Hence,

$$\begin{gathered} g_{gre}=0.05+2\times 0.02 \\ =0.09 \end{gathered}$$