Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
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Answer:
The wave with 0.03 time period have the greater frequency
Explanation:
Given in the question,
Period T of one wave = 0.03 seconds
Period T of another wave = 1/4 seconds
We know that
T = 1 / f
So,
f = 33.3 hz
f = 4 hz
The wave with 0.03 time period have the greater frequency.
Answer:
- The difference in length for steel is 2.46 x 10⁻⁴ m
- The difference in length for invar is 1.845 x 10⁻⁵ m
Explanation:
Given;
original length of steel, L₁ = 1.00 m
original length of invar, L₁ = 1.00 m
coefficients of volume expansion for steel, = 3.6 × 10⁻⁵ /°C
coefficients of volume expansion for invar, = 2.7 × 10⁻⁶ /°C
temperature rise in both meter stick, θ = 20.5°C
Difference in length, can be calculated as:
L₂ = L₁ (1 + αθ)
L₂ = L₁ + L₁αθ
L₂ - L₁ = L₁αθ
ΔL = L₁αθ
Where;
ΔL is difference in length
α is linear expansivity =
Difference in length, for steel at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m
Difference in length, for invar at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m