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....
