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

The answer is "1.278498"

Step-by-step explanation:

Given formula:

`f = ((1)/(2L)) \sqrt{((T)/(m))}`

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

`\to (2380)/(2237) = ((L + x))/(L)\\\\\to (2380)/(2237) = 1 + ((x)/(L))\\\\\to 1.0639249=1+((x)/(L))\\\\\to 1.0639249 -1=((x)/(L))\\\\\to (x)/(L)= 0.0639249\\\\ \to x= 0.0639249* 20.0 \\\\ \to x= 1.278498`