Question

A wire of length L is made up of two sections of two different materials connected in series. The first section of length L1 = 17.7 m is made of steel and the second section of length L2 = 28.5 m is made of iron. Both wires have the same radius of 5.30 ✕ 10−4 m. If the compound wire is subjected to a tension of 148 N, determine the time taken for a transverse pulse to move from one end of the wire to the other. The density of steel is 7.75 ✕ 103 kg/m3 and the density of iron is 7.86 ✕ 103 kg/m3.

Answer 1

Answer:

Explanation:

velocity of wave in a tense wire is given by the expression

$v= \sqrt{\frac{T}{m} }$

v is velocity . T is tension and m is mass per unit length .

for steel wire

m = π r² ρ where r is radius and ρ is density

= 3.14 x (5.3 x 10⁻⁴)²x7.75 x 10³

= 683.57 x 10⁻⁵ kg/m

v =  $\sqrt{\frac{148}{683.57\times 10^{-5}} }$

= 1.47 x 10² m /s

= 147 m /s

for iron  wire

m = π r² ρ where r is radius and ρ is density

= 3.14 x (5.3 x 10⁻⁴)²x7.86 x 10³

= 693.27 x 10⁻⁵ kg/m

$v = \sqrt{\frac{148}{693.27\times 10^{-5}} }$

= 146 m /s

Time taken to move from one end to another

= 17.7 / 147 + 28.5 / 146

= .12 + .195

= .315 s .