Having difficulties with pendulum theory and percentage error homework problem.

Question

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.

Answer 1

There are a lot of ways to propagate error of

$$ T = 2\pi \sqrt{\dfrac{L}{g}} $$

1. You can use a numeric, like

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

2. You can approximate the expression using a plane

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

3. 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

$$T \pm \Delta T = 2\pi \sqrt{\dfrac{L(1\pm \Delta x)}{g(1\pm \Delta y)}}$$ $$T \pm \Delta T = T \cdot \left(1\pm \Delta x\right)^{1/2} \cdot \left(1\pm \Delta y\right)^{-1/2}$$ $$ T \pm \Delta T = T \cdot \left(1 \pm \dfrac{1}{2} \Delta x\right) \cdot \left(1 \mp \dfrac{1}{2} \Delta y\right)$$ $$ \Delta T = T \cdot \left(\dfrac{1}{2} \Delta x + \dfrac{1}{2} \Delta y\right)$$