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Two rods A and B are connected end to end. Sound wave travelling in rod A has a speed . If itenters the rod B whose Young’s modulus is four times that of A but having the same density as A,the speed of the sound wavee in rod B will be

Question

Two rods A and B are connected end to end. Sound wave travelling in rod A has a speed . If itenters the rod B whose Young’s modulus is four times that of A but having the same density as A,the speed of the sound wavee in rod B will be

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Solution

The speed of sound in a rod (or any medium) is given by the formula:

v = sqrt(Y/ρ)

where: v is the speed of sound, Y is the Young's modulus of the material, and ρ is the density of the material.

Given that the Young's modulus of rod B is four times that of rod A, and the density is the same, we can write the speed of sound in rod B as:

v_B = sqrt(4Y_A/ρ_A) = 2*sqrt(Y_A/ρ_A) = 2v_A

So, the speed of sound in rod B will be twice the speed of sound in rod A.

This problem has been solved

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