Tension of String Question?

Question

A violin string is 36.0 long. It sounds the musical note A (440 Hz) when played without fingering.
How far from the end of the string should you place your finger to play the note C (523 Hz)?

Any help on how to approach this question would be appreciated. Thank you

Ansr is 5.71cm

Answer 1

Use the old v=fλ rule, where:

v = speed of sound
f1 = frequency of the first note
λ1 = wavelength of the first note
etc.

v = fλ ──► f = v/λ

f2 / f1 = (v/λ2) / (v/λ1) = λ1 / λ2
λ2 = λ1(f1 / f2) = 36(440 / 523) = 30.3 cm

This is how far the OTHER end of the string the finger should be placed (i.e. from the bridge). How far from the neckstop is just the difference of the two wavelengths:
Δλ = λ2 - λ1 = 36 - 30.3 = 5.7 cm

Answer 2

T*sin theta = m*(w^2)*r -------------------------- a million and T*cos theta = m*g --------------------------------- 2 the place theta = semi-vertical attitude of the cone which the string describes = tan^-a million[30/40] = tan^-a million[3/4] and w is the angular velocity of the around action of the conical pendulum Dividing a million by utilising 2, we get tan theta = 3/4 = [(w^2)*r]/g or w^2 = 3*g/(4*r). So a million may be rewritten as T*sin theta = m*(w^2)*r= (3*m*g)/4 Squaring and including a million and a pair of we get T^2 = [(mg)^2]*[(3/4)^2 + a million^2] or T = 0.a million*9.8sq rt[9/sixteen +a million] = (5*0.a million*9.8)/4 = a million.225 N

Answer 3

fundamental frequency
f= v/ 2L

For a string of tention T, mass m and length L velocity is given as
v= sqrt(T/(m/L))

then f= sqrt(T/(m/L))/2L

f= sqrt( T/(4m L)) where
T- string tension
m =mass of the string
L- string length

We cam say that L =k/ (f )^2 where (k=T/4m) and can be found from a earlier frequency
k=Lf^2
L'= L(f/f')^2= 36 (440/523)^2=25.48 cm

Answer 4

Just use proportions.

The frequency of a note is inversely proportionate to the length of the string. (longer strings play lower notes).


so f1 = alpha/L1 assuming alpha is the same then

f2 = alpha/L2 so

f1*L1 = f2*L2 solving for L2 we get

L2 = f1/f2*L1 = 30.2868

then the distance from the end is L1 - L2 = 5.7132