A violin has an open string length (bridge to nut) of L=32.7 cm. At what distance x from the bridge does a violinist have to place his finger on the fingerboard to play a C (523.3 Hz) on the A string (fundamental frequency 440 Hz)?
To determine the distance x from the bridge where the violinist should place his finger to play a C (523.3 Hz) on the A string (fundamental frequency 440 Hz), we can use the formula for the frequency of a vibrating string:
f = (1 / 2L) * sqrt(T / μ)
where:
- f is the frequency of the played note (in Hz)
- L is the open string length (bridge to nut)
- T is the tension in the string (in N)
- μ is the linear mass density of the string (in kg/m)
In this case, we know the fundamental frequency of the A string is 440 Hz. We can use this information to find the value of T. The relationship between the fundamental frequency and the tension is given by:
f1 = (1 / 2L) * sqrt(T / μ)
440 = (1 / (2 * 0.327)) * sqrt(T / μ) [Substituting L = 32.7 cm = 0.327 m]
Solving for T / μ:
T / μ = (440 * 2 * 0.327)^2 = 1217.4
Now that we have T / μ, we can calculate the frequency f for the note C (523.3 Hz):
523.3 = (1 / (2 * L)) * sqrt(T / μ) [Substituting f = 523.3 Hz]
Solving for x, we need to rearrange the formula:
x = L - ((v / f) * L)
where:
- x is the distance from the bridge where the finger should be placed
- L is the open string length (bridge to nut)
- v is the speed of sound in air (approximately 343 m/s at room temperature)
Substituting the known values, we have:
x = 0.327 - ((343 / 523.3) * 0.327)
Calculating this expression:
x ≈ 0.174 meters
Therefore, the violinist needs to place his finger approximately 17.4 cm (or 0.174 m) from the bridge to play a C (523.3 Hz) on the A string.
To find the distance x from the bridge where the violinist needs to place his finger on the fingerboard to play a C (523.3 Hz) on the A string (fundamental frequency 440 Hz), we can use the equation:
f = (v/2L) * √(1 + (x/L))
Where:
f = frequency of the desired note (523.3 Hz)
v = speed of sound (approximately 340 m/s)
L = open string length (32.7 cm = 0.327 m)
x = distance from the bridge to the finger placement point
First, let's convert the open string length from centimeters to meters:
L = 0.327 m
Next, we can rearrange the equation and solve for x:
x = L * (√((f/(v/2L))^2 - 1))
Plugging in the values:
x = 0.327 * (√((523.3 / (340 / (2*0.327)))^2 - 1))
Now, let's calculate:
x ≈ 0.327 * (√((523.3 / 1.0475)^2 - 1))
x ≈ 0.327 * (√((499769.759)^2 - 1))
x ≈ 0.327 * (√(249769362359.28 - 1))
x ≈ 0.327 * (√(249769362358.28))
x ≈ 0.327 * 499768.9162
x ≈ 163426.0469 cm
Therefore, the violinist needs to place his finger approximately 163426.0469 cm from the bridge to play a C (523.3 Hz) on the A string.
fundamental frequency (ff) = 440 Hz
frequency at C (fc)= 523.3 Hz
L = 32.7 cm
L(C) = ?
x(C) = ?
L (C) = L * ff/fc
L (C) = 27.50 cm
x (C) = L - L(C) = 5.20 cm
Finger has to be placed at 5.20 cm from the end.