Two wires are anchored at both ends. Wire 1 has a mass of 105 grams and a length of 6.9 m. Wire 2 has a mass of 60 grams, a length of 20 m, and a tension of 143 N. What must be the tension in Wire 1 if its third harmonic is to equal the first overtone of Wire 2?
T_{wire1} =
YOu are given mass/length. You are given half wavlength length of 6.9m, and 20 m.
wire 1 has a third harmonic wavelength of 30 m. First overtone of wire2 has a wavlength of 2*6.9m
but what is the tension of the wire
To find the tension in Wire 1 (T_wire1) when its third harmonic is equal to the first overtone of Wire 2, we can use the formula for the frequency of a stretched string:
f = (1/2L) * √((T/μ))
where f is the frequency, L is the length of the wire, T is the tension, and μ is the linear mass density.
First, let's find the linear mass density (μ) for both wires:
For Wire 1:
μ_1 = m_1 / L_1
where m_1 is the mass of Wire 1 and L_1 is the length of Wire 1.
For Wire 2:
μ_2 = m_2 / L_2
where m_2 is the mass of Wire 2 and L_2 is the length of Wire 2.
Next, we can equate the frequencies of the third harmonic of Wire 1 and the first overtone of Wire 2:
f_1 = 3f_2
where f_1 is the frequency of Wire 1 (third harmonic) and f_2 is the frequency of Wire 2 (first overtone).
Now, let's substitute the formulas for frequency and linear mass density into the equation:
(1/2L_1) * √((T_1/μ_1)) = 3 * (1/2L_2) * √((T_2/μ_2))
Substituting the known values:
(1/2 * 6.9) * √((T_1 / (m_1 / L_1))) = 3 * (1/2 * 20) * √((143 / (m_2 / L_2)))
Simplifying the equation:
√((T_1 / (m_1 / L_1))) = 3 * √((143 / (m_2 / L_2)))
Now, we can solve for T_wire1 by isolating it on one side of the equation:
T_1 = (m_1 * L_1 * 3 * √((143 / (m_2 * L_2)))) / 6.9
Plugging in the given values:
T_1 = (105 * 6.9 * 3 * √((143 / (60 * 20)))) / 6.9
Simplifying further:
T_1 ≈ 106.44 N
Therefore, the tension in Wire 1 (T_wire1) must be approximately 106.44 N for its third harmonic to equal the first overtone of Wire 2.