Three metal rods are located relative to each other as shown in the figure below

Question

Three metal rods are located relative to each other as shown in the figure below, where L3 = L1 + L2. The speed of sound in a rod is given by

Y/? Three metal rods are located relative to each other as shown in the figure below, where L3 = L1 + L2. The speed of sound in a rod is given by where Y is Young's modulus for the rod and ? is the density. Values of density and Young's modulus for the three materials are ?1 = 2.70 times 103 kg/m3, Y1 = 7.00 times 1010 N/m2, ?2 = 1.20 times 104 kg/m3,Y2 = 1.60 times 1010 N/m2, ?3 = 8.80 times 103 kg/m3, Y3 = 1.14 times 1011 N/m2. If L3 = 1.50 m, what must the ratio L1/L2 be if a sound wave is to travel the length of rods 1 and 2 in the same time interval required for the wave to travel the length of rod 3?

Explanation / Answer

v1 = sqrt(Y1/p1) = 5.091 x 10^3 m/s

v2 = sqrt(Y2/p2) = 1.155 x 10^3 m/s

v3 = sqrt(Y3/p3) = 3.599 x 10^3 m/s

Now,

v1 = L1/t1 , v2 = L2/t2 and v3 = L3/t3

Given,

t1 + t2 = t3

L1/v1 + L2/v2 = L3/v3

So,

L1v2 + L2v1 = L3*v1*v2/v3

1155L1 + 5091L2 = 2450.72

But, L1 + L2 = 1.5

So, L2 = 1.5 - L1

1155L1 + 1.5*5091 - 5091L1 = 2450.72

3936L1 = 5185.78

L1 = 1.318 m L2 =0.182 m

So, L1/L2 = 7.22 = 7

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