Question

A violin string which is 30.4 cm long is tuned so that its pitch is concert A, which is 440 Hz. I can effectively shorten the string by placing my finger on it and pressing down. How far (cm) from the end of the string should I put my finger if I want the pitch of the note to be 539 Hz? (Give the distance from the end where the wooden tuning pegs are located.) Assume that the tension in the string remains the same. Give the answer to 2 significant figures.

Answer 1

Answer: 24.82 cm from the end.

Explanation: In order to explain this problem we have to consider the relationship for a fundamenal wave in a string, which  is given by:

f= v/2L where f is the frequency; v the speed of the wave and L the length of the string.

for f=440 Hz we can calculate the speed which is equal:

v=f*2*L  the speed depen d the linear mass of the string and the tension we keep the same for the new frequency (539 Hz), so  the equation is now:

fnew=v/2*Lnew, so

Lnew=v/2*fnew =f*2*L/(2*fnew)=(f/fnew)*L then we have:

Lnew=(440/539)*30.4=28.82 cm

Answer 2

Answer:

The distance from the end of the rope where you must put your finger is 5.58 cm

Explanation:

The relation between tension, mass, frecuency and length is inversely related. The expression is equal:

$f_{1} L_{1} =f_{2} L_{2} \\L_{2} =\frac{f_{1}L_{1} }{f_{2} }$

Where

f₁ = 440 Hz

L₁ = 30.4 cm

f₂ = 539 Hz

Replacing:

$L_{2} =\frac{440*30.4}{539} =24.82cm$

The distance from the end of the rope where you must put your finger is:

L = 30.4 - 24.82 = 5.58 cm