Question:What must be the stress (F/A) in a stretched wire of a material whose Young’s modulus is Yfor the speed of longitudinal waves to equal 30 times the speed of transverse waves?

The stress is given as, F/A.

Young modulus is given as, Y.

When a sound wave is traveling in the medium in such a way that the vibration of the particles of the medium is parallel to the direction of propagation of the wave, then it is known as a longitudinal wave. If the particles of the medium are vibration perpendicular to the direction of the propagation of the wave, the wave is known as a transverse wave.

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.