Unit 6 - Problem 5 ==> Electropendulum

Given:

• $$L = 0.75 \;\mbox{meters}$$
• $$m = 0.02 \;\mbox{kg}$$
• $$E = 10 \;\mbox{n/s}$$
• $$q = 0.01 \;\mbox{coulombs}$$

Rationale:

• $$T = 2 \pi \sqrt\frac{L}{a}$$
• $$F = m * a$$
• $$a = \frac{F}{m}$$  
• $$T = 2 \pi \sqrt\frac{L}{\frac{F}{m} } = 2 \pi \sqrt\frac{(m * L)}{F}$$
• $$F_e = E * q$$
• $$T =  2 \pi \sqrt\frac{(m * L)}{(E * q)}$$

Calculate

• $$T =  2 \pi \sqrt\frac{(0.02 * 0.75)}{(10.0 * 0.01)}$$  
• $$T =  2 \pi \sqrt\frac{0.015}{ 0.1}$$
• $$T =  2 \pi \sqrt{0.15}$$  
• $$T =  2 \pi * 0.387298$$  
• $$T = 2.43 \;\mbox{seconds}$$