The string on a violin has a fundamental frequency of 120.0 Hz. The length of the vibrating portion is 36 cm and has a mass of 0.58 g. Under what tension (in newtons) must the string be placed?
The fundamental frequency is: f1 = 120Hz
At the fundamental, the string has two nodes at the end, and one anti-node in the middle, so therefore the vibrating length of the string (0.36 m) is equal to λ/2 , so λ= 0.72 m.
v = λ•f = 0.72• 120 = 86.4 m/s.
Now we will apply the wave speed in a string relationship, which is:
v = sqrt(T/mₒ),
where mₒ = m/L= 0.58•10^-3/0.36 = 1.6•10^-3 kg/m is mass/length for the cord,
and T is the tension in the cord.
Therefore, T = mₒ•v² =1.6•10^-3 •(86.4)² ≈ 12 N.
To find the tension in the string, we can use the formula for calculating the fundamental frequency of a string:
f = (1/2L) * sqrt(T/μ)
where f is the fundamental frequency, L is the length of the vibrating portion of the string, T is the tension in the string, and μ is the linear mass density of the string.
We can rearrange this formula to solve for the tension:
T = (4L^2 * f^2 * μ)
Given:
f = 120.0 Hz
L = 36 cm = 0.36 m
μ = mass/length = 0.58 g / 0.36 m = 1.61 g/m = 0.00161 kg/m
Substituting these values into the equation:
T = (4 * (0.36 m)^2 * (120.0 Hz)^2 * 0.00161 kg/m)
Now, let's calculate the tension:
T = (4 * 0.1296 m^2 * 14400.0 Hz^2 * 0.00161 kg/m)
T = (0.0939924 kg * m * Hz^2)
To convert the unit, we need to multiply it by the gravitational force constant (g):
T = (0.0939924 kg * m * Hz^2 * 9.8 m/s^2)
T = 0.92124552 N
Therefore, the tension in the string must be approximately 0.9212 N.