Question

Standing wave produced in a metal rod of length $1m$ is represented by the equation $y={10}^{-6}\sin \frac{\pi x}{2}\sin 200\pi t$ where $x$ is in metre and $t$ is in seconds. The maximum tensile stress at the mid-point of the rod is (Young's modulus of material of rod $={10}^{12}N/m^2$ ).

• A. $\frac{\pi }{2}\times {10}^{6}N/m^2$
• B. $2\pi \times {10}^{6}N/m^2$
• C. $\frac{\pi }{2\sqrt{2}}\times {10}^{6}N/m^2$
• D. $\frac{2\pi }{\sqrt{3}}\times {10}^{6}N/m^2$

Answer 1

Answer: (C)

$\frac{\pi }{2\sqrt{2}}\times {10}^{6}N/m^2$

The tensile stress is given by,
Tensile stress = Y $\times$ tensile strain
= Y $\times \left(\frac{dy}{dx}\right)$
Since, Y is a material constant,
For maximum tensile stress, tensile strain should be maximum

strain , $\frac{dy}{dx}=\frac{\pi }{2}{10}^{-6}\cos \frac{\pi x}{2}\sin 200\pi t$

at mid point , x = L/2 and $\sin 200\pi t=1$ (to maximize), maximum strain would be,

$\frac{\pi }{8^{0.5}}\times {10}^{-6}$

Multiplying this with Y, will give us the maximum tensile stress.
Thus, maximum tensile stress is = $\frac{\pi }{8^{0.5}}\times {10}^6$