Question

A simple pendulum with a period of 2 seconds can be used as a chronograph to measure time with great accuracy. A platinum wire attached to the bob completes the electric timing circuit through a drop of mercury as it swings though the lowest point. a. What should be the length, L, of the pendulum? b. If the platinum wire is in contact with the mercury for 0.317 cm of the swing, what must be the initial 1. amplitude to limit the duration of contact to 0.01 seconds? (Assume that the velocity during contact is constant and that the amplitude of oscillation is small.) 508 cm-0.317cm

Answer 1

Explanation:

Answer:

0.994 m

Explanation:

For a simple pendulum, the period of oscillation is given by

T=2\pi\dfrac{\sqrt{L}}{g}T=2π

g

L

wher L is its length and g s the acceleration due to gravity.

Rearranging the formula,

L=\left g(\dfrac{T}{2\pi}\right)^2

T = 2

Take g = 9.81

L=\left 9.81(\dfrac{2}{2\pi}\right)^2

L=\left 9.81(\dfrac{1}{\pi}\right)^2=0.994