Answer to Question #220292 in Mechanics | Relativity for Lesega

Question

A 3.00 m long steel wire is fixed in a ceiling and a mass of 12.3 kg is hanging from the wire. The wire has a diameter d = 0.80 mm, a Young’s modulus 2.00 GPa and mass density 7860 kg/m3 .

(a) Calculate the stress in the wire.

(b) Calculate the strain.

(c) Calculate the elongation of the wire.

(d) Calculate the transverse speed of pulse.

(e) Calculate the longitudinal sound speed.

Answer 1

Explanations & Calculations

• Assume the thread's mass to be negligible compared to the hanging mass hence a light string.
• Load applied to the wire = $\small mg = 12kg\times9.8ms^{-2} = 120.54\,N$
• Area of the wire = $\small \pi \frac{d^2}{4} = \pi \times \frac{(0.80\times10^{-3}m)^2}{4}=5.03\times10^{-7}m^2$

(a)

• Strain is given by

$\qquad\qquad \begin{aligned} \small \sigma&=\small \frac{F}{A}\\ &=\small \frac{120.54\,N}{5.03\times10^{-7}m^2}\\ &=\small 2.396\times10^8\\ & \approx\small 2.40\times10^8\,Pa \end{aligned}$

(b)

• Using the stress-strain equation, the strain can be found

$\qquad\qquad \begin{aligned} \small \sigma&=\small Y\epsilon\\ \small \epsilon&=\small \frac{2.40\times10^8Pa}{2.00\times10^9Pa}\\ &=\small \bold{0.12} \end{aligned}$

(c)

• Once the strain is known, elongation can be calculated

$\qquad\qquad \begin{aligned} \small \epsilon&=\small \frac{e}{L}\\ \small e&=\small 0.12\times3.00m\\ &=\small \bold{0.36 m} \end{aligned}$

(d)

• Transverse speed is given by

$\qquad\qquad \begin{aligned} \small v&=\small \sqrt{\frac{T}{\mu}}=\sqrt{\frac{F}{\mu}}=\sqrt{\frac{F}{A\rho}}\\ &=\small \sqrt{\frac{120.54N}{5.03\times10^{-7}m^2\times7860kgm^{-3}}}\\ &=\small \bold{174.61\,ms^{-1}} \end{aligned}$

(e)

• Normally longitudinal speeds are described for stiff rod-like objects, and this thread could be thought similar to such a situation as it has some mass(own mass) & thickness.
• Longitudinal speed is given by

$\qquad\qquad \begin{aligned} \small v_L&=\small \sqrt{\frac{E}{\rho}}\\ &=\small \sqrt{\frac{2.00\times10^9Pa}{7860kgm^{-3}}}\\ &=\small \bold{504.43\,ms^{-1}} \end{aligned}$