An object of mass $$m$$ is suspended at the end of a massless wire of length $$L$$ and area of cross-section, A. Young modulus of the material of the wire is $$Y$$. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:

We consider a wire of equilibrium length $$L$$, cross-sectional area $$A$$ and Young modulus $$Y$$. A mass $$m$$ is hanging at its lower end. First, recall the definition of Young’s modulus:

$$Y=\frac{\text{stress}}{\text{strain}} =\frac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}} =\frac{F\,L}{A\,\Delta L}.$$

Here $$F$$ is the tensile force producing an extension $$\Delta L$$. Solving the above relation for the extension produced by a force $$F$$, we get

$$\Delta L=\frac{F\,L}{Y\,A}.$$

Now imagine that the suspended mass is displaced very slightly downward so that the wire stretches an extra small amount $$x$$ (in addition to whatever static extension already exists). The corresponding extra restoring force $$F_x$$ that develops in the wire is obtained by replacing $$\Delta L$$ with this incremental stretch $$x$$:

$$x=\frac{F_x\,L}{Y\,A}\quad\Longrightarrow\quad F_x=\frac{Y\,A}{L}\,x.$$

We observe that this relation has the standard Hooke’s-law form $$F_x=kx$$, where the effective spring constant $$k$$ of the wire is

$$k=\frac{Y\,A}{L}.$$

Once the wire behaves like a spring of constant $$k$$, the system becomes a simple vertical mass-spring oscillator. The angular frequency $$\omega$$ of such an oscillator is given by the well-known formula

$$\omega=\sqrt{\frac{k}{m}}.$$

Substituting the value of $$k$$ we have just obtained,

$$\omega=\sqrt{\frac{Y\,A/L}{m}} =\sqrt{\frac{Y\,A}{m\,L}}.$$

The ordinary frequency $$f$$ (number of oscillations per second) is related to the angular frequency by $$f=\dfrac{\omega}{2\pi}$$. Therefore,

$$f=\frac{1}{2\pi}\sqrt{\frac{Y\,A}{m\,L}}.$$

This expression matches option B.

Hence, the correct answer is Option B.