Question

In a damped oscillator with $m = 500g$, $K = 100N/m$ and $b = 75g/s$, the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of the 20 cycles is given as ${e^{ - x}}$. Find x.

Answer 1

Hint:In order to solve this numerical problem we need to find the time period of the damped oscillator in a single cycle and then multiply it by the number of cycles. Then use the expression to find the initial amplitude.

Formula Used:The time period of a damped harmonic oscillator is given by the expression mentioned below:
$T = 2\pi \sqrt {\dfrac{m}{K}} $
Here m is the mass of the damped harmonic oscillator and K is the kinetic energy of the damped harmonic oscillator.

Step by step answer:
The mass of the oscillator is given as $m = 500g = 0.5kg.$ Also; the kinetic energy of the oscillator is given as $K = 100N/m.$ now putting these respective values in the formula for obtaining the time period, we have:
$T = 2\pi \sqrt {\dfrac{{0.5}}{{100}}} = \dfrac{\pi }{{5\sqrt 2 }}$
This time period obtained is only for one complete cycle. We need to find the time period of the damped harmonic oscillator for 20 complete cycles as given in the numerical problem. Thus,
$t = 20 \times \dfrac{\pi }{{5\sqrt 2 }} = 2\sqrt 2 \pi $
Now, the ratio of amplitude after 20 complete cycles to the initial amplitude of the damped harmonic oscillator is given by the mathematical expression given below:
$\dfrac{y}{{{y_\circ }}} = {e^{ - bt}}$
Now the value of b from the numerical problem is given as $b = 75g{s^{ - 1}}$. We have to convert this value to its SI unit. So, we have:
$b = \dfrac{{75}}{{1000}}kg/s$
We can now substitute the values of the respective variables in the above equation. Then we get:
$\dfrac{y}{{{y_ \circ }}} = {e^{\left( {\dfrac{{ - 75 \times 2\sqrt 2 \pi }}{{1000}}} \right)}} = {e^{ - 0.666}}$
Thus, we find that the value of x is -0.666 from the above expression.

Note:It is important to convert every expression to its respective SI unit as the equality will not be dimensionally correct otherwise.