High School

A loudspeaker is found to set a violin string placed near it into oscillation at frequencies 966.0 Hz and 1449.0 Hz only. This loudspeaker is fed by an audio oscillator with a frequency range of 800 Hz to 1500 Hz. If the violin string is 31.0 cm long with a linear density of 0.718 g/m, calculate the tension in the string.

Answer :

Final answer:

The tension in a violin string that oscillates at 966.0 Hz, we can use the formula for the frequency of a standing wave on a string, rearranged to solve for tension. Assuming that 966.0 Hz is the fundamental frequency, we can calculate the tension by substituting the given values of string length and linear density into the formula.

Explanation:

To calculate the tension in a violin string that oscillates at given frequencies when exposed to sound from a loudspeaker, we can use the formula for the frequency of a standing wave on a string: f = (n/2L) * sqrt(T/μ), where f is the frequency, n is the harmonic number, L is the length of the string, T is the tension, and μ is the linear density (mass per length).


Since the frequencies mentioned are 966.0 Hz and 1449.0 Hz, these are likely to represent different harmonics of the string. The fundamental frequency is typically the lowest frequency at which the string oscillates, which would be when n=1. However, since we're not given which harmonics these frequencies correspond to, we will assume that 966.0 Hz is the fundamental frequency (n=1).



We can rearrange the formula to solve for T: T = (2Lf)^2 μ. Given that L is 31.0 cm (or 0.310 m) and μ is 0.718 g/m (or 0.000718 kg/m), we can substitute these values into the formula: T = (2 * 0.310 m * 966.0 Hz)^2 * 0.000718 kg/m. Computing this will yield the tension in the string in newtons.