What is the wavelength of the third harmonic for a 1.24-m length of 18-gauge copper wire with a 160.0-N ball hanging from it? How does the wavelength change when a 500.0 N ball replaces the original one?
Wavelength of Third Harmonic
The wavelength of the third harmonic for the copper wire is calculated by finding the fundamental frequency first and then determining the wavelength based on the harmonic number.
Using the given data:
Length of wire (L) = 1.24 m
Weight of ball (m) = 160.0 N
Diameter of wire (d) = 1.024 mm
Density of copper (ρ) = 8.96 x 10^3 kg/m^3
Speed of sound in copper (v) = 4.7 km/s = 4.7 x 10^3 m/s
First, we need to calculate the linear mass density (μ) of the wire using the formula:
μ = (π/4) * d^2 * ρ
μ = (π/4) * (1.024 x 10^-3 m)^2 * 8.96 x 10^3 kg/m^3
μ = 2.944 x 10^-7 kg/m
Next, we find the tension in the wire (T) by considering the weight of the ball and the supporting force:
T = mg + F
F = μgL
F = 2.944 x 10^-7 kg/m * 9.8 m/s^2 * 1.24 m
F = 3.615 x 10^-4 N
T = 160.0 N + 3.615 x 10^-4 N
T = 160.0 N
Now, we can calculate the fundamental frequency:
f1 = (v/2L) * sqrt(T/μ)
f1 = (4.7 x 10^3 m/s) / (2 * 1.24 m) * sqrt(160.0 N / 2.944 x 10^-7 kg/m)
f1 = 3.005 x 10^3 Hz
The wavelength of the third harmonic is three times the length of the wire, so:
λ3 = 3 * 1.24 m
λ3 = 3.72 m
Therefore, the wavelength of the third harmonic for the wire is 3.72 meters.
Change in Wavelength with Heavy Ball
When the 500.0 N ball replaces the 160.0 N one, the tension in the wire increases to T = 500.0 N.
Calculating the new fundamental frequency:
f1' = (v/2L) * sqrt(T/μ)
f1' = (4.7 x 10^3 m/s) / (2 * 1.24 m) * sqrt(500.0 N / 2.944 x 10^-7 kg/m)
f1' = 3.667 x 10^3 Hz
The frequency of the third harmonic with the heavy ball is:
f3' = 3f1'
f3' = 1.100 x 10^4 Hz
The wavelength of the third harmonic with the heavy ball remains the same as before:
λ3' = 2 * 1.24 m / 3
λ3' = 0.827 m
The change in the wavelength caused by replacing the light ball with the heavy one is calculated as:
Δλ3 = (λ3' - λ3) / λ3 * 100%
Δλ3 = (0.827 m - 3.72 m) / 3.72 m * 100%
Δλ3 = -0.048%
This means that the wavelength of the third harmonic decreases by approximately 0.048%.
In summary, the wavelength of the third harmonic for the copper wire is initially 3.72 meters. When the heavy ball replaces the light one, the wavelength decreases by approximately 0.048%.