The transmissibility ratio will be equal to unity for all values of damping factor, if \(\dfrac{\omega }{{{\omega _n}}}\;\) is equal to:

Explanation:

In forced vibration, the transmissibility ratio is given by,

\(\Rightarrow TR = \dfrac{{{F_T}}}{{{F_0}}} = \dfrac{{\sqrt {1 + {{\left( {2\xi q} \right)}^2}} }}{{\sqrt {{{\left( {1 - {q^2}} \right)}^2} + {{\left( {2\xi q} \right)}^2}} }}\)

Where,

\(q = {\bf{Frequency}}\;{\bf{ratio}} = \dfrac{\omega }{{{\omega _n}}}\)

• When, q = 0 ⇒ TR = 1, (independent of ζ)
• When, q = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When, q = √2, ⇒ All the curves pass through the point TR = 1 for all values of damping factor ξ.
• When q < √2 ⇒ TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When q > √2 ⇒ TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus, vibration isolation is possible only in the range of q > √2. Here the force transmitted to the foundation increases as the damping is increased.