Question

A pendulum with a length of 1.00 m is released from an initial angle of 15.0° . After 1000s , its amplitude has been reduced by friction to 5.50° . What is the value of b/2 m ?

Answer 1

A pendulum with a length of 1.00 m is released from an initial angle of 15.0° and after 1000s , its amplitude has been reduced to 5.50°, where the value of the $\frac{b}{2m}$ is $0.001$ $sec^{-1}$.

The type of oscillation shown here is a damped oscillation, where slowly with respect to time, the amplitude of oscillation decreases.

In the damped oscillation, the position is determined according to the equation given below:

$x(t)=Ae^{-\frac{bt}{2m} }$

The given information:

$l= 1.00 m\\x(0)=15^{o} \\x(1000)=5.50^{o}$

So using the equation:

$x(t)=Ae^{-\frac{bt}{2m} }$

$\frac{x(1000)}{x(0)} =\frac{Ae^{-\frac{b(1000)}{2m} }}{Ae^{-\frac{b(0)}{2m} }} \\\frac{x(1000)}{x(0)} =\frac{Ae^{-\frac{b(1000)}{2m} }}{Ae^{-\frac{b(0)}{2m} }} =\frac{5.50}{15} \\\therefore e^{-\frac{b(1000)}{2m} } = =\frac{5.50}{15} \\\therefore \frac{b}{2m} = 0.001 \ s^{-1}$

Hence, the value of the $\frac{b}{2m}$ is $0.001$ $sec^{-1}$.

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