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Compare the oscillations in an LC circuit are analogous to the oscillation of a block at the end of a spring. |
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Answer» Solution :The LC oscillation is similar to the mechanical oscillation of a block attached to a spring. The LOWER part of each figure in question no. 22 depicts the corresponding stage of a mechanical system. Differential EQUATION of LC oscillation, `(d^(2)q)/( dt^(2))+( q)/( LC) =0` or ` (d^(2)q)/(dt^(2)) + omega_(0)^(2)q = 0` for a block of mass m oscillating with frequency `omega_(0)` the equation is, `(d^(2)x)/(dt^(2))+omega_(0)^(2) = 0` or ` (d^(2) x)/( dt^(2)) + ( Kx)/( m ) = 0 ` Here `omega_(0) = sqrt((k)/( m))` and k is the spring constant. From F = kc in mechanics `k = ( F )/( x ) ` means the external force requires to produce unit extension or compression is known as force constant. Unit of force constant `NM^(-1)`. Corresponding equation of this equation in LC circuit is` V = (q)/( C )` `:. (1)/(C ) = ( V )/(q)` means for established unit charge,necessary electrical difference is `(1)/( C )`. In below table analogies between mechnical and electrical quantities are shown. `{:("Mechanical system","Electrical system"),("Mass m","Inductance L"),("Force constant k","Reciprocal capacitance "1//C),("Displacement x","Charge q"),("Velocity" v=(dx)/(dt),"Current" I = ( dq)/(dt)),("Mechanical energy" E = (1)/(2) kx^(2) + ( 1)/(2) m v^(2),"Electromagnetic energy" E = ( 1)/(2) (q^(2))/(C ) kx^(2) + ( 1)/(2) m v^(2) ):}` |
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