Question

A 2.00-m long piano wire with a mass per unit length of 12.0 g/m is under a tension of 8.00 kn. What is the frequency of the fundamental mode of vibration of this wire?

Answer 1

Answer:

204.1 Hz

Explanation:

The fundamental frequency of a vibrating string is given by:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the linear mass density of the string

For the wire in this problem, we have

L = 2.00 m

T = 8.00 kN = 8000 N

$\mu = 12.0 g/m = 0.012 kg/m$

Therefore, substituting into the equation, we find the frequency of the string:

$f=\frac{1}{2(2.00 m)}\sqrt{\frac{8000 N}{0.012 kg/m}}=204.1 Hz$