Answer :
The highest string on a violin is tuned to E^5. The vibrating part of the string on a full-size violin is approximately 32.7 cm long.
The highest string on a violin is tuned to E^5. This means it produces the note E in the fifth octave of the musical scale. Each octave corresponds to a doubling of frequency. The fifth octave of the E note has a frequency of approximately 659.255 Hz.
To calculate the length of the vibrating part of the string on a full-size violin, we can use the formula for the fundamental frequency of a vibrating string:
f = (1/2L)√(T/μ)
Where:
f = frequency of the string,
L = length of the vibrating part of the string,
T = tension in the string, and
μ = linear density of the string.
For a full-size violin, the tension is typically around 60 N, and the linear density is approximately 0.02 kg/m. Substituting these values into the formula and rearranging to solve for L, we get:
L = (1/2) × (T / (f^2 × μ))
L = (1/2) × (60 / ((659.255)^2 × 0.02))
L ≈ 0.327 m
Converting meters to centimeters, the length of the vibrating part of the string is approximately 32.7 cm. Therefore, the vibrating part of the highest string on a full-size violin is about 32.7 cm long.
