Question

A guitar string is \(90\,\text{cm}\) long and has a fundamental frequency of \(124\,\text{Hz}\). Where should it be pressed to produce a fundamental frequency of \(186\,\text{Hz}\)?

• A. \(20\,\text{cm}\) from an end
• B. \(40\,\text{cm}\) from an end
• C. \(50\,\text{cm}\) from an end
• D. \(60\,\text{cm}\) from an end

Hint:

Use \(f\propto1/L\): find the new sounding length, then locate the fret.

Answer 1

The Correct Option is D

Step 1: For a stretched string the fundamental frequency is \[f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\] The tension and mass per unit length do not change when you press the string, so \(f\) is inversely proportional to the vibrating length \(L\): \[f\propto\frac{1}{L}\quad\Rightarrow\quad f_1L_1=f_2L_2\]
Step 2: Find the new vibrating length: \[L_2=L_1\cdot\frac{f_1}{f_2}=90\times\frac{124}{186}=90\times\frac{2}{3}=60\,\text{cm}\]
Step 3: So only \(60\,\text{cm}\) of the string must vibrate. When you press the string on a fret, the part between the pressed point and the bridge vibrates. To leave a \(60\,\text{cm}\) vibrating segment, the finger is placed \(60\,\text{cm}\) from that end (equivalently \(30\,\text{cm}\) from the other end).
Step 4: Therefore the string is pressed \(60\,\text{cm}\) from an end. \[\boxed{60\,\text{cm from an end}}\]