Question

The energy in an oscillating LCcircuit containing an$\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{\text{H}}$inductor is$\mathbf{5}\mathbf{.}\mathbf{70}\mathbf{}\mathit{\mu }\mathit{J}$. The maximum charge on the capacitor is$\mathbf{175}\mathbf{}\mathit{\mu }\mathit{C}$. For a mechanical system with the same period, (a)Find the mass. (b) Find the spring constant. (c) Find the maximum displacement. (d) Find the maximum speed.

Short answer

1. The mass of the mechanical system is $1.25\text{kg}$.
2. The spring constant of the mechanical system is $372N/m$.
3. The maximum displacement of the system is $1.75\times {10}^{-4}\text{m}$.
4. The maximum speed of the system is $3.02\times {10}^{-3}\text{m/s}$.

Step-by-step solution

The given data

1. Inductance of the inductor in the LC circuit ,$L=1.25\text{H}$
2. Energy in LC circuit,role="math" localid="1662813087505" $U=5.70\times {10}^{-6}\text{J}$
3. Maximum charge on capacitor,$q=175\times {10}^{-6}\text{C}$

Understanding the concept of analogy between the LC system and the mechanical block system

In the given LC circuit, the energy stored by the inductor and capacitor is similar to the concept of a mechanical system. Thus, using the analogy between the LC system and block spring system, we will find the mass, spring constant, maximum displacement, and maximum velocity of the mechanical system.

Formulae:

The electric energy stored by the capacitor in an LC circuit, $\mathit{E}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{2}\mathbf{C}}$ (i)

The magnetic energy stored by the inductor in the LC circuit, $\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{L}{\mathit{i}}^{\mathbf{2}}$ (ii)

The current amplitude of the LC circuit, $\mathit{i}\mathbf{=}\mathit{q}\mathit{\omega }$ (iii)

The potential energy of a spring-block system, $\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$ (iv)

The kinetic energy of a spring-block system, $\mathit{K}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$ (v)

The angular frequency of the LC circuit, $\mathit{\omega }\mathbf{=}\frac{\mathbf{1}}{\mathbf{\surd }\mathbf{L}\mathbf{C}}$ (vi)

a) Calculation of the mass of the system

Before starting calculations, we compare the formulae from equations (i), (ii), (iv) and (v) in LC circuit and spring block system. Thus, the analogy between the energies of the LC circuit and the spring-block system gives us the following relations:

Spring constant,$k=\frac{1}{C}$

Maximum displacement,$x=q$

Mass $m=L$, and

Maximum velocity,$v=i$

As explained above, mass in mechanical system isthesame as inductance L in LC circuit. That is:

$$\begin{gathered} m=L \\ =1.25\text{kg} \end{gathered}$$

Hence, the value of the mass is $1.25\text{kg}$.

b) Calculation of the spring constant

From part (a) calculations, we can say that the spring constant in mechanical system is equivalent to reciprocal of capacitance in LC circuit.

$k=\frac{1}{C}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(1\right)$

To find capacitance, we will use electric energy in LC circuit that is given by equation (i).

Thus,$U=E$

Rearranging the equation for capacitance C, we can get the capacitance value from above equation as follows:

$$\begin{gathered} C=\frac{q^2}{2U} \\ =\frac{{(175\times {10}^{-6})}^2}{2\times 5.70\times {10}^{-6}} \\ =2686\times {10}^{-6}\text{F} \end{gathered}$$

Using in equation 1, we get the spring constant of the system as follows:

$$\begin{gathered} k=\frac{1}{2686\times {10}^{-6}} \\ =372\text{N/m} \end{gathered}$$

Hence, the spring constant of mechanical system is $372\text{N/m}$.

c) Calculation of the maximum displacement

From part (a) calculations, the maximum displacement corresponds to the maximum charge in LC circuit. That is:

$$\begin{gathered} x=q \\ =1.75\times {10}^{-4}\text{m} \end{gathered}$$

Hence, the value of the maximum displacement is $1.75\times {10}^{-4}\text{m}$.

d) Calculation of the maximum velocity

From part (a) calculations, the maximum velocity corresponds to the current in LC circuit.

Using Angular frequency equation (vi) of LC system in equation (iii), we can get the current value as follows:

$$\begin{gathered} i=\frac{q}{\sqrt{LC}} \\ =\frac{175\times {10}^{-6}}{\surd 1.25\times 2686\times {10}^{-6}} \\ =3.02\times {10}^{-3}\text{A} \end{gathered}$$

Thus, the value of the maximum velocity through analogy is given by,$v_{max}=3.02\times {10}^{-3}\text{m/s}$

Hence, the maximum velocity is $3.02\times {10}^{-3}\text{m/s}$.