The non-zero value of ω, for which the amplitude of the force transmitted to the ground will be F0, is

The correct option is $\sqrt{\frac{2k}{m}}$.

Understanding the System from the Image:
The schematic diagram in the first image shows a single-degree-of-freedom mass-spring-damper system consisting of:
• A mass labeled $m$
• A spring of stiffness coefficient $k$ (labeled $K$ in the second image)
• A damper with damping coefficient $c$ (labeled $C$ in the second image)
• A harmonic excitation force acting on the mass, given by:
$$F\left(t\right)=F_0\cos \left(\omega t\right)$$
where $F_0$ is the amplitude of the excitation force, and $\omega$ is the excitation frequency.

Transmissibility Ratio:
The force transmitted to the ground support, $F_T$, is carried through both the spring and the damper. The ratio of the amplitude of this transmitted force to the excitation force amplitude is defined as the force transmissibility ratio ($ϵ$ or $\mathrm{TR}$):
$$ϵ=\frac{F_T}{F_0}$$
We are given that the amplitude of the transmitted force is exactly equal to the excitation amplitude, meaning:
$$F_T=F_0\Rightarrow ϵ=1$$

Mathematical Derivation:
The formula for force transmissibility in a damped system is:
$$ϵ=\frac{\sqrt{1+{\left(2\zeta r\right)}^2}}{\sqrt{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}}$$
where:
• $r=\frac{\omega }{{\omega }_n}$ is the frequency ratio.
• ${\omega }_n=\sqrt{\frac{k}{m}}$ is the natural frequency of the system.
• $\zeta$ is the damping ratio.

Setting $ϵ=1$:
$$1=\frac{\sqrt{1+{\left(2\zeta r\right)}^2}}{\sqrt{{(1-r^2)}^2+{\left(2\zeta r\right)}^2}}$$
Squaring both sides and simplifying:
$$1+{\left(2\zeta r\right)}^2={(1-r^2)}^2+{\left(2\zeta r\right)}^2$$
Subtracting ${\left(2\zeta r\right)}^2$ from both sides yields:
$${(1-r^2)}^2=1$$
Taking the square root of both sides:
$$1-r^2=\pm 1$$

This gives us two cases:
1. Case 1 (taking the positive sign):
$$1-r^2=1\Rightarrow r^2=0\Rightarrow \omega =0$$
Since the problem asks for the non-zero value of $\omega$, we discard this solution.

2. Case 2 (taking the negative sign):
$$1-r^2=-1\Rightarrow r^2=2$$
Taking the positive square root for physical frequency:
$$r=\sqrt{2}$$

Substituting back $r=\frac{\omega }{{\omega }_n}$:
$$\frac{\omega }{{\omega }_n}=\sqrt{2}\Rightarrow \omega =\sqrt{2}{\omega }_n$$
Since the natural frequency is defined as ${\omega }_n=\sqrt{\frac{k}{m}}$, we have:
$$\omega =\sqrt{2}\sqrt{\frac{k}{m}}=\sqrt{\frac{2k}{m}}$$