A string A has twice the length, twice the diameter, twice the tension and twice the density of another string B. The overtone of A which will, have the same fundamental frequency as that of B is

Question

TestHub · Accepted Answer

The fundamental frequency of a stretched string is given by $f = \frac{1}{2 L} \frac{T}{μ}$ , where $μ$ is the linear mass density.

Since $μ = ρ \times π (\frac{d}{2})^{2} = \frac{π ρ d ^{2}}{4}$ , the frequency can be written as:

$f = \frac{1}{L d} \frac{T}{π ρ}$

For string B, let the fundamental frequency be $f_{B} = \frac{1}{L d} \frac{T}{π ρ}$ .

For string A, we are given $L_{A} = 2 L$ , $d_{A} = 2 d$ , $T_{A} = 2 T$ , and $ρ_{A} = 2 ρ$ .

The fundamental frequency of string A is:

$f_{A} = \frac{1}{( 2 L ) ( 2 d )} \frac{2 T}{π ( 2 ρ )} = \frac{1}{4 L d} \frac{T}{π ρ} = \frac{1}{4} f_{B}$

We need the $n$ -th harmonic of A to have the same frequency as the fundamental frequency of B:

$n f_{A} = f_{B} \Rightarrow n (\frac{1}{4} f_{B}) = f_{B} \Rightarrow n = 4$

For a string fixed at both ends, the $n$ -th harmonic corresponds to the $(n - 1)$ -th overtone. Therefore, the 4th harmonic is the 3rd overtone.

Answer: third

Physics - Wave on String Question with Solution | TestHub

Options:

Doubts & Discussion