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. $f=\left(\frac{1}{2\pi }\right)\surd \left(\frac{YA}{mL}\right)$
• B. $f=\left(\frac{1}{2\pi }\right)\surd \left(\frac{mL}{YA}\right)$
• C. $f=\left(\frac{1}{2\pi }\right)\surd \left(\frac{YL}{mA}\right)$
• D. $f=\left(\frac{1}{2\pi }\right)\surd \left(\frac{mA}{YL}\right)$

Answer 1

The correct option is A

$f=\left(\frac{1}{2\pi }\right)\surd \left(\frac{YA}{mL}\right)$

Step 1. Given data :

Mass $m$,

Length $L$,

Area of cross section $A$,

Youngs modulus $Y$,

Step 2. Find the frequency :

Young's modulus is

$Y=\frac{FL}{A∆L}$ (Where,$F$ is force, $∆L$ is change in length).

$F=\left(\frac{AY}{L}\right)∆L$…….(1)

The force in terms of the spring constant is,

$F=K∆L$……..(2)

From the equation (1) and (2) of force, the spring constant $K$ is,

$K=\frac{YA}{L}$

Frequency , $f=(\frac{1}{2\pi })\surd \left(\frac{K}{m}\right)$

$f=\left(\frac{1}{2\pi }\right)\surd \left(\frac{YA}{mL}\right)$

Hence, option (A) is correct .