A damped harmonic oscillator consists of a block ($\mathbf{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$), a spring ($\mathbf{k}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$), and a damping force ($\mathbf{F}\mathbf{=}\mathbf{-}\mathbf{bv}$). Initially, it oscillates with amplitude of$\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$; because of the damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. (a) What is the value of b? (b) How much energy has been “lost” during these four oscillations?

• The spring constant is, $k=10\mathrm{N}/\mathrm{m}$.
• Mass of the block is$M=2.00\mathrm{kg}$.
• The amplitude of oscillation is,$x_m=25\mathrm{cm}\mathrm{or}0.25\mathrm{m}$.
• Because of damping, $x=\frac{3}{4}{x}_m$.

We can find the damping constant using the formula for the displacement of the damped oscillator and period of SHM. Using this constant and damping amplitude, we can find the energy at t=4T. From this, we can find the energy lost during 4 oscillations.

Formulae:

The displacement equation of the damped oscillations,$x=x_m{e}^{-\frac{bt}{2m}}$(i)

The period of oscillation for SHM,$T=2\pi \sqrt{\frac{m}{k}}$(ii)

The potential energy of the system, $\left(PE\right)=\frac{1}{2}k{x}^2$(iii)

Using the displacement equation (i), we get

$e^{-\frac{bt}{2m}}=\frac{3}{4}(\because x=\frac{3}{4}{x}_m)$

Since,$t=4T$,

$e^{-\frac{b\left(4T\right)}{2m}}=\frac{3}{4}$(iv)

The period of oscillation for SHM using equation (ii)is given as:

$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{2}{10}} \\ =2.81\mathrm{s} \end{gathered}$$

Then, substituting the value of time period in equation (iv), we get

$$\begin{gathered} e^{\mathit{-}\frac{\mathit{b}\mathit{(}\mathit{4}\mathit{\times }\mathit{2}\mathit{.}\mathit{81}\mathit{)}}{\mathit{2}\mathit{m}}}=\frac{3}{4} \\ \frac{b(4\times 2.81)}{2m}=In\left(\frac{4}{3}\right) \\ \mathrm{b}=\frac{In\left({\displaystyle \frac{4}{3}}\right)\left(2m\right)}{11.24} \\ =\frac{0.288\left(2\right)\left(2\right)}{11.24} \\ =0.102\mathrm{kg}/\mathrm{s} \end{gathered}$$

Therefore, the value of b is 0.102 kg/s .

The initial energy of the system is,

$$\begin{gathered} E=P.E \\ =\frac{1}{2}k{x^2}_m \\ =\frac{1}{2}(10.0\mathrm{N}/\mathrm{m}){(0.250\mathrm{m})}^2=0.313\mathrm{J} \end{gathered}$$

At role="math" localid="1657268926058" $\mathrm{t}=4\mathrm{T}$,the energy of the damped system is given as;

$$\begin{gathered} E\text{'}=\frac{1}{2}k{\left(\frac{3}{4}{x}_m\right)}^2(\because x=\frac{3}{4}{x}_m) \\ =\frac{9}{16}(0.313)\mathrm{J} \\ =0.176\mathrm{J} \end{gathered}$$

The energy lost during 4 oscillations is,

$$\begin{gathered} E-E\text{'}=0.313\mathrm{J}-0.176\mathrm{J} \\ =0.137\mathrm{J} \end{gathered}$$

Therefore, the energy lost during 4 oscillations is 0.137 J .