Question

Estimate frequency of vibration of your arm. Let the length of the arm be 0.57 m. Consider the arm as a simple pendulum and assume its center of mass is about 2/3 its length from the shoulder. Express your answer with the appropriate units. Estimate the natural period of vibration of your arm.

Answer 1

Answer:

0.80865 Hz

1.23662 seconds

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

l = Length of arm = 0.57 m

Length of simple pendulum is given by

$L=\dfrac{2}{3}l\\\Rightarrow L=\dfrac{2}{3}\times 0.57\\\Rightarrow L=0.38\ m$

The frequency is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.81}{0.38}}\\\Rightarrow f=0.80865\ Hz$

The frequency is 0.80865 Hz

The time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{0.80865}\\\Rightarrow T=1.23662\ s$

The time period is 1.23662 seconds