The velocity of the longitudinal wave can be written as,
\({v_{{\rm{longitudinal}}}} = \sqrt {\frac{Y}{\rho }} \)
Here\(\rho \)is the density of the medium.
The velocity of the transverse wave can be written as,
\({v_{{\rm{transverse}}}} = \sqrt {\frac{F}{\mu }} \)
Here\(\mu \)is the mass per unit length.
We need to find the stress when the velocity of the longitudinal wave is equal to 30 times the velocity of the transverse wave.
Thus, we can write,
\({v_{{\rm{longitudinal}}}} = 30{v_{{\rm{transverse}}}}\)
Substitute the values in the above expression, and we get,
\(\begin{array}{c}\sqrt {\frac{Y}{\rho }} = 30\sqrt {\frac{F}{\mu }} \\\frac{Y}{\rho } = 900\frac{F}{\mu }\end{array}\) (1)
The density and mass per unit length are related as,
\(\frac{\rho }{A} = \mu \)
Where\(A\)is the cross-section area of the wire.
Now equation 1 can be written as,
\(\begin{array}{c}\frac{Y}{\rho } = 900\frac{F}{{\frac{\rho }{A}}}\\Y = 900\frac{F}{A}\\\frac{F}{A} = Y/900\end{array}\)
Thus, stress must be \(\frac{F}{A} = Y/900\), which means 1/900 of the young modulus of the wire to satisfy the given condition.
