Question

A segment A of wire stretched tightly between two posts a distance L apart vibrates in its fundamental mode with frequency f. A segment B of an identical wire is stretched with the same tension, but between two different posts. You observe that the frequency of the second harmonic of wire B is the same as the fundamental frequency of wire A. The length of wire B must be
A) ½L.
B) L.
C) 2L.
D) 4L.

Answer 1

Answer:

option (c)

Explanation:

Fundamental frequency of segment A = f

Second harmonic frequency of B = fundamental frequency of A .

Tension in both the wires is same and the mass density is also same as the wires are identical.

fundamental frequency of wire A is given by

$f=\frac{1}{2L_{A}}{\sqrt{\frac{T}{m}}}$    .... (1)

Second harmonic of B is given by

$f=\frac{2}{2L_{B}}{\sqrt{\frac{T}{m}}}$    .... (2)

Equation (1) is equal to equation (2), we get

$\frac{1}{2L_{A}}=\frac{2}{2L_{B}}$

$L_{B}=2L_{A}$

So, LB = 2 L

Thus, the length of wire segment B is 2 times the length of wire segment A.