9. Figure shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between
two fixed supports. The tension produced is 40 N. The cross-sectional area of the steei wire is 1.0 mm2 and
that of the aluminium wire is 3.0 mm2. The minimum frequency of a tuning fork which can produce standing
waves in the system with the joint as a node is 10P (in Hz) then find P. Given density of aluminium is
2.6 g/cm3 and that of steel is 7.8 g/cm3.
Hint:
Stress in steel = 40N/1mm^2
Stress in aluminum = 40N/3mm^2
Speed of sound = (stress/density)^1/2
Wavelength × Frequency = Speed
Minimum frequency corresponds to a half wavelength in one of the two wires. Find out which wire and use the corresponding frequency.
To find the value of P, we need to use the equation for the minimum frequency of a tuning fork that can produce standing waves in a stretched string or wire. The equation is given by:
f = (1/2L) * √(T/μ)
Where:
- f is the frequency in Hz
- L is the length of the wire
- T is the tension in the wire
- μ is the linear mass density of the wire
Let's calculate the μ for both the steel and aluminum wires first:
- For the steel wire:
The cross-sectional area of the steel wire is given as 1.0 mm² (or 0.01 cm²). The density of steel is given as 7.8 g/cm³.
The linear mass density (μ) of the steel wire can be calculated as:
μ = (mass / length) = (density * volume / length)
Since the volume = area * length, we can substitute it in the equation:
μ = (density * area) = (7.8 g/cm³ * 0.01 cm²) = 0.078 g/cm
- For the aluminum wire:
The cross-sectional area of the aluminum wire is given as 3.0 mm² (or 0.03 cm²). The density of aluminum is given as 2.6 g/cm³.
The linear mass density (μ) of the aluminum wire can be calculated as:
μ = (mass / length) = (density * volume / length)
Since the volume = area * length, we can substitute it in the equation:
μ = (density * area) = (2.6 g/cm³ * 0.03 cm²) = 0.078 g/cm
Now that we have the values of μ for both wires, we can substitute them into the equation for the minimum frequency:
f = (1/2L) * √(T/μ)
Given:
- Length of the aluminum wire = 60 cm
- Length of the steel wire = 80 cm
- Tension = 40 N
Let's calculate the frequency:
For the aluminum wire:
f₁ = (1/2(60 cm)) * √(40 N / 0.078 g/cm) ≈ 16.72 Hz
For the steel wire:
f₂ = (1/2(80 cm)) * √(40 N / 0.078 g/cm) ≈ 12.54 Hz
Since the minimum frequency of a tuning fork that can produce standing waves in the system with the joint as a node is given by:
f = f₁ - f₂
Substituting the values:
10P = f₁ - f₂
Now, we can solve for P:
P = (f₁ - f₂) / 10
Substituting the calculated values of f₁ and f₂:
P = (16.72 Hz - 12.54 Hz) / 10 = 0.418 Hz
Therefore, P is approximately equal to 0.418 Hz.