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?

Answer 1

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.