A string with a mass/length of 0.00200 kg/m is stretched between two points with a tension of 120 N. If the fundamental frequency is 440 Hz, then what is the distance between the two points?
Question
A string with a mass/length of 0.00200 kg/m is stretched between two points with a tension of 120 N. If the fundamental frequency is 440 Hz, then what is the distance between the two points?
Solution
To solve this problem, we need to use the formula for the fundamental frequency of a vibrating string, which is given by:
f = (1/2L) * sqrt(T/μ)
where: f is the frequency, L is the length of the string, T is the tension in the string, and μ is the linear mass density of the string.
We are given that f = 440 Hz, T = 120 N, and μ = 0.00200 kg/m. We need to solve for L.
Rearranging the formula to solve for L gives us:
L = sqrt(T/μ) / (2f)
Substituting the given values into this equation gives us:
L = sqrt(120 N / 0.00200 kg/m) / (2 * 440 Hz)
Calculating the square root first:
sqrt(120 N / 0.00200 kg/m) = sqrt(60000 m²/s²) = 245.0 m/s
Then dividing by 2f:
L = 245.0 m/s / (2 * 440 Hz) = 0.278 m
So, the distance between the two points (i.e., the length of the string) is approximately 0.278 meters.
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