Having difficulties with pendulum theory and percentage error homework problem. The pendulum theory：
$$t=2 \pi \sqrt{l/g},$$
where

*

*$t$ is the time of period,

*$L$ is the length of the pendulum,

*$G$ is the acceleration due to the gravity (~9.81 m/s²).

Calculate the expected percentage error in the time period if the measurement of length is 5% high.
My lectures says you need to do this by binomial expansion, but I don't know how to do that.
 A: There are a lot of ways to propagate error of
$$
T = 2\pi \sqrt{\dfrac{L}{g}}
$$

*

*You can use a numeric, like


$$
T_{med} = 2\pi \sqrt{\dfrac{L_{med}}{g_{med}}}
$$
$$
T_{max} = \max \left(2\pi \sqrt{\dfrac{L_{med}\pm \Delta L}{g_{med} \pm \Delta g}}\right)
$$
$$
T_{min} = \min\left(2\pi \sqrt{\dfrac{L_{med}\pm \Delta L}{g_{med} \pm \Delta g}}\right)
$$
$$
\Delta T = \dfrac{1}{2}\left(T_{max}-T_{min}\right)
$$



*You can approximate the expression using a plane


$$
\Delta T = \left|\dfrac{\partial T}{\partial L}\right| \cdot \Delta L + \left|\dfrac{\partial T}{\partial g}\right| \cdot \Delta g
$$
$$ \Delta T = \dfrac{\pi \Delta L}{\sqrt{Lg}} + \pi \Delta g \sqrt{\dfrac{L}{g^3}}$$
$$ \Delta T = T \cdot \left[\dfrac{\Delta L}{2L}+\dfrac{\Delta g}{2g}\right]$$



*Using binomial/taylor expansion like


Treating $g$ as constant, then
$$
T \pm \Delta T = 2\pi \sqrt{\dfrac{L\pm \Delta L}{g}} = 2\pi \sqrt{\dfrac{L}{g}\left(1\pm \Delta x\right)}
$$
Where $\Delta x = \Delta L/L$
$$
T \pm \Delta T = \underbrace{2\pi \sqrt{\dfrac{L}{g}}}_{T} \left(1\pm \Delta x\right)^{1/2}
$$
Doing the expansion of $(1\pm \Delta x)^{n}$ we have
$$
(1\pm \Delta x)^{n} \approx 1 \pm n\Delta x + \dfrac{n(n-1)}{2}\Delta x^2 + \Theta(\Delta x^3)
$$
If $\Delta x$ is small, then we can suppress the term $\Delta x^2$ and get
$$ T \pm \Delta T = T \left(1\pm \Delta x\right)^{1/2} \approx T \left(1+\dfrac{1}{2} \cdot \Delta x\right)$$
$$\boxed{\Delta T = T \cdot \dfrac{1}{2}\Delta x}$$
If $g$ changes, you can use this response and get
$$T \pm \Delta T = 2\pi \sqrt{\dfrac{L(1\pm \Delta x)}{g(1\pm \Delta y)}}$$
$$T \pm \Delta T = T \cdot \left(1\pm \Delta x\right)^{1/2} \cdot \left(1\pm \Delta y\right)^{-1/2}$$
$$ T \pm \Delta T = T \cdot \left(1 \pm \dfrac{1}{2} \Delta x\right) \cdot \left(1 \mp \dfrac{1}{2} \Delta y\right)$$
$$ \Delta T = T \cdot \left(\dfrac{1}{2} \Delta x + \dfrac{1}{2} \Delta y\right)$$

