Question

When the thinnest string is pressed down at the fret closest to the bridge (d = 20.0 cm) and the string plucked near the midpoint, the vibrating string produces a sound with a frequency of 2,380 Hz. If the guitarist moves to the next fret (second-closest to the bridge), the sound produced has a frequency of 2,237 Hz. Determine the distance (in cm) between the frets. (Assume the given frequencies are fundamental frequencies.)

Answer 1

Answer:

The answer is "1.278498"

Explanation:

Given formula:

$f = (\frac{1}{2L}) \sqrt{(\frac{T}{m})}$

$f ( \ at\ L = 20.0) \ = 2380\\\\for \ L + x ,\\\\ f = 2237\ H_zso,$

$\to \frac{2380}{2237} = \frac{(L + x)}{L}\\\\\to \frac{2380}{2237} = 1 + (\frac{x}{L})\\\\\to 1.0639249=1+(\frac{x}{L})\\\\\to 1.0639249 -1=(\frac{x}{L})\\\\\to \frac{x}{L}= 0.0639249\\\\ \to x= 0.0639249\times 20.0 \\\\ \to x= 1.278498$