A steel wire in a piano has a length of 0.6000 m and a mass of 3.000 10-3 kg. To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C (fC = 261.6 Hz on the chromatic musical scale)?
To find the tension in the wire, we need to use the wave equation for a vibrating string, which relates the tension, mass, length, and frequency of the wave.
The wave equation for a vibrating string is:
v = √(T/μ)
Where:
v is the wave velocity (speed of propagation of the wave in the string),
T is the tension in the string,
μ is the linear mass density of the string (mass per unit length).
The linear mass density (μ) is calculated by dividing the mass of the wire (m) by its length (L):
μ = m/L
Given:
Length of the wire (L) = 0.6000 m
Mass of the wire (m) = 3.000 x 10^-3 kg
Frequency of middle C (fC) = 261.6 Hz
First, find the linear mass density (μ):
μ = m/L
= (3.000 x 10^-3 kg) / (0.6000 m)
= 5.000 x 10^-3 kg/m
Now, we need to find the wave velocity (v) using the formula v = √(T/μ). Rearranging the formula, we have:
T = v^2 * μ
Since the frequency of the fundamental vibration (f) is related to the wave velocity (v) and the length (L) of the wire by the equation:
v = f * λ
where λ is the wavelength
For a fixed wave velocity and frequency, the wavelength is inversely proportional to the frequency (λ = v/f). Thus:
λ = v / f
λ = L / n (where n is the number of nodes)
Since the fundamental frequency has one node (n = 1), the wavelength is equal to twice the length of the wire (λ = 2L).
Substituting the values, we get:
2L = v / f
v = 2L * f
Now, we can find the wave velocity (v):
v = 2L * f
= 2 * 0.6000 m * 261.6 Hz
≈ 314.0 m/s
Finally, substitute the values of v and μ into the equation T = v^2 * μ to find the tension (T):
T = v^2 * μ
= (314.0 m/s)^2 * 5.000 x 10^-3 kg/m
≈ 5.000 x 10^2 N
Therefore, the tension in the wire must be approximately 500 N for the fundamental vibration to correspond to middle C (261.6 Hz) on the chromatic musical scale.