Question

A rod of length one meter is rigidly clamped at its midpoint. Longitudinal stationary waves are set up in such a manner that there are two nodes on either side from its midpoint. The amplitude of anti-node is 2 micrometer. What is the frequency of the rod if Young's modulus of the rod is $2\times {10}^{11}N/m^2$ and the density of rod is $8000kg/m^3$?

• A. $12500Hz$
• B. $6250Hz$
• C. $25000Hz$
• D. $3125Hz$

Answer 1

Answer: (A)

$12500Hz$

Velocity of sound $v=\sqrt{\frac{Y}{\rho }}=\sqrt{\frac{2\times {10}^{11}}{8\times {10}^3}}=5000m/s$

There are two nodes on either side of the clamped midpoint.

The distance between two consecutive nodes is $\frac{\lambda }{2}$ and

distance between a node and antinode is $\frac{\lambda }{4}$

Hence, $4\times \left(\frac{\lambda }{2}\right)+2\times \left(\frac{\lambda }{4}\right)=L$

$\Rightarrow \frac{5}{2}\lambda =L$

$\Rightarrow \lambda ==\frac{2}{5}\times 1=\frac{2}{5}=0.4m$

$\therefore \nu =\frac{v}{\lambda }=\frac{5000}{0.4}=12500Hz$