Question

A steel wire of diameter 2.000 mm and length 1.000 m is attached between two immovable supports.When the temperature is 60.00 Celsius the tension in the wire is (essentially) zero. The temperature is then reduced to 20.00 Celsius. Young's Modulus for steel is 2.000E10 Pa and the density of steel is 7.860E3 kg/m3 and the coefficient of linear expansion of steel is 11.0E-6 C-1.
A. Find the tension in the steel wire.
B. What would the velocity of a traveling wave be at this lower temperature?
C. Find the pattern of allowed frequencies of standing waves allowed on this wire.

Answer 1

Answer:

a. 27.65 N b. 5.93 cm/s c. 0.03n where n = 1,2,3....

Explanation:

a. Since Young's Modulus E = stress/strain = σ/ε and σ = εE, we find the stain in the steel wire by finding its strain at 20°C.

Now Δl = lαΔθ where l = length of steel wire = 1.000 m, α = coefficient of linear expansion of steel = 11.0 × 10⁻⁶ C⁻¹ and Δθ = change in temperature = 20 °C - 60 °C = -40 °C

The strain  ε = Δl/l = αΔθ = 11.0 × 10⁻⁶ C⁻¹ × -40 °C = -4.4 × 10⁻⁴

σ = T/A= εE, where T = tension in steel wire and A = cross-sectional area of steel wire

T = εEA = εEπd²/4 where d = diameter of wire = 2.000 mm = 2 × 10⁻³ m

So,

T = εEπd²/4

= 4.4 × 10⁻⁴ × 2× 10¹⁰ Pa × π ×  (2 × 10⁻³ m )²/4

= 27.65 N

b. The velocity of the travelling wave at this lower temperature is

v = √(T/μ) where μ = mass per unit length = ρl where ρ = density of steel wire = 7.86 × 10³ kg/m³ and l = length of wire = 1.000 m

v = √(T/μ)

= √(T/ρl)

= √(27.65 N/(7.86 × 10³ kg/m³ × 1.000 m))

= √(27.65 N/7.86 × 10³ kg/m²)

= √(0.3517 × 10⁻² N/kg/m²)

= 0.05931 m/s

≅ 0.0593 m/s

= 5.93 cm/s

c. To find the allowable frequencies, we first find the fundamental frequency f₀ = v/2l

= 0.0593 m/s ÷ (2 × 1 m)

= 0.0593 m/s ÷ 2

= 0.0297 Hz

≅ 0.03 Hz

So, the allowable modes of frequency are multiples of f₀, that is fₙ = nf₀ where n = 1,2,3....

So, fₙ = 0.03n Hz where n = 1,2,3....