Question

Two wires are made of the same material but the second wire has twice the diameter and twice the length of the first wire. When the two wires are stretched, and the tension in the second wire is also twice the tension in the first wire, the fundamental frequency of the first wire is 330 Hz. What is the fundamental frequency of the second wire? Answer in units of Hz.

Answer 1

Answer:

116.7 Hz

Explanation:

Let there are two wires A and B.

Tension in wire A = T

Tension in wire B = 2T

Length of wire A = L

Length of wire B = 2L

fundamental frequency in wire A, fA = 330 Hz

let the fundamental frequency in wire B is fB.

The formula for the fundamental frequency is given by

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

where, μ is the mass per unit length

mass per unit length of wire A = Area of wire A x density

mass per unit length of wire B = Area of wire B x density

$\frac{\mu _{A}}{\mu _{B}}=\frac{d_{A}^{2}}{4d_{A}^{2}}=\frac{1}{4}$

So,

$\frac{f_{A}}{f_{B}}=\frac{L_{B}}{L_{A}}{\sqrt{\frac{T_{A}\mu _{B}}{T_{B}\mu _{A}}}}$

$f_{B}=\frac{f_{A}}{2\sqrt{2}}$

fB = 330 / 2.828

fB = 116.7 Hz

Thus, the frequency in the second wire is 116.7 Hz.