Question

The amplitude of a lightly damped oscillator decreases by 3.0 %during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

Short answer

The percentage of mechanical energy lost in each cycle is 6%.

Step-by-step solution

The given data:

The amplitude of the oscillator decreases by 3% in each cycle.

Understanding the concept of damping:

Damping is the restrain of the vibratory motion of a body affecting its oscillations due to the dissipation of energy. Here, the oscillator is damped resulting in a constant decrease of the energy that results in restricting its motion that is a decrease in its amplitude.

Formula:

The mechanical energy of the body undergoing oscillations,

$\mathrm{E}=\frac{1}{2}{\mathrm{kx}}_{\mathrm{m}}^2$ ….. (i)

Where, k is the spring constant and x_m is the amplitude or the maximum displacement of the body.

Calculate percentage of the mechanical energy of the oscillator is lost in each cycle:

Let the initial mechanical energy of the oscillator according to equation (i) be,

$\mathrm{E}=\frac{1}{2}{\mathrm{kx}}_{\mathrm{m}}^2$ ….. (I)

Here, x_m is the initial amplitude of the oscillator.

You have the amplitude of the oscillations decreased by 3% .

So, the amplitude of next oscillation will become,

data-custom-editor="chemistry" $\begin{array}{rcl}{{\mathrm{x}}_{\mathrm{m}}}^{\text{'}}& =& {\mathrm{x}}_{\mathrm{m}}-0.03{\mathrm{x}}_{\mathrm{m}}\\ & =& 0.97{\mathrm{x}}_{\mathrm{m}}\end{array}$

Thus, the decreased mechanical energy of the next cycle of oscillator is given using the above data in equation (i) as follows:

$\begin{array}{rcl}{\mathrm{E}}^{\text{'}}& =& \frac{1}{2}\mathrm{k}{\left({{\mathrm{x}}_{\mathrm{m}}}^{\text{'}}\right)}^2\\ & =& \frac{1}{2}\mathrm{k}{\left(0.97{\mathrm{x}}_{\mathrm{m}}\right)}^2\\ & =& \left(0.94\right)\frac{1}{2}\mathrm{k}{\left({{\mathrm{x}}_{\mathrm{m}}}^{\text{'}}\right)}^2\\ & =& \left(0.94\right)\mathrm{E}\end{array}$

So, the loss in the amount of mechanical energy of the oscillator is given as follows:

$\begin{array}{rcl}∆\mathrm{E}& =& {\mathrm{E}}^{\text{'}}-\mathrm{E}\\ & =& 0.94\mathrm{E}-\mathrm{E}\\ & =& -0.06\mathrm{E}\\ \frac{∆\mathrm{E}}{\mathrm{E}}& =& -0.06\\ & =& 6\%\end{array}$

The negative sign shows that energy is lost.

Hence, the percentage of mechanical energy lost in each cycle is 6%.