Animated Solution for Physics - Oscillations: 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

Imagine a heavy block of mass $m$ hanging peacefully from a thin wire. The wire has a length $L$, a cross-sectional area $A$, and is made of a material characterized by its Young's modulus $Y$. When we pull the mass down slightly and let go, it starts oscillating up and down. At first glance, this looks like a complex elasticity problem. But there is a beautiful, simplifying trick we can use.

An elastic wire behaves exactly like a spring! When you stretch it, the intermolecular forces act like tiny springs pulling it back to its original length. We can model this entire setup as a simple spring-mass system.

For a wire, the restoring force comes from its elasticity. According to the definition of Young's modulus:

Rearranging this to solve for the restoring force $F$, we get:

Notice how this perfectly mirrors Hooke's Law for a spring, $F=Kx$, where $x$ is the extension $\Delta L$. By comparing the two, we can immediately see that the equivalent spring constant $K$ of our wire is:

Now that we have our equivalent spring constant, the rest is straightforward kinematics. We know the time period $T$ of a standard spring-mass system is given by:

Let's substitute our equivalent spring constant $K=\frac{YA}{L}$ into this formula:

The question, however, asks for the frequency of oscillation. Frequency $f$ is simply the reciprocal of the time period ($f=1/T$). Flipping our expression, we get:

And there we have it! The frequency depends directly on the stiffness of the material ($Y$) and its thickness ($A$), and inversely on the suspended mass ($m$) and the length of the wire ($L$). This perfectly matches option (b).