Question

A copper

Answer 1

Answer:

v = 64.14 m/s

Explanation:

First we are going to calculate the initial tension force in the wire.

How:

$V = \sqrt{\frac{T}{\mu} }$

      Where  v: Wave's speed

                   T. Tension force

                   μ: Linear densityof the wire

Then:

$T = V^{2}\mu$

$T = 46^{2}(7 x 10^{-3}) = 0.322N$

Now we calculate the  linear dilation of the copper, thus

ΔL = αLoΔT

  Where α:The coefficient of linear expansion for copper

ΔL = $(17x10^{-6})( Lo)(-14) = 0.000238 Lo$

The Young's modulus is defined like:

$E = \frac{TLo}{A\Delta L}$

        Where E : The Young's modulus

                    A: cross-sectional area

Then

$T = \frac{EA\Delta L}{Lo}$

$T = \frac{1.1 x10^{11} (1.1x10^{-6} )(0.000238Lo)}{Lo} =28.798 N$

and the speed of the wave when the temperature is lowered is:

$v = \sqrt{\frac{28.798}{7x10^{-3} } } = 64.14 m/s$