Question

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:

• A. $f=\frac{1}{2\pi }\sqrt{\frac{mL}{YA}}$
• B. $f=\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}$
• C. $f=\frac{1}{2\pi }\sqrt{\frac{mA}{YL}}$
• D. $f=\frac{1}{2\pi }\sqrt{\frac{YL}{mA}}$

Answer 1

The correct option is B $f=\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}$

An elastic wire can be treated as a spring and its spring constant is given as
$k=\frac{YA}{L}[\because Y=\frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta l}{l_0}\right)}]$
Thus, frequency of oscillation will be
$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}=\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}$
Hence, $\left(B\right)$ is the right option.