Question

Two strings have the same length and tension. One string has a mass per length that is 4 times that of the other string. The fundamental frequency of the more massive string will be times _________ (a) larger, (b) smaller than that of the less massive string.

Answer 1

Answer:

$f_1=\frac{f_2}{2}$

Explanation:

the Frequency of string is given by

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

T= tension in the string

μ= mass per unit length of string

l= length of string

in the given question $\mu_{1}$= 4$\mu_{2}$

here subscript 1 is massive string and subscript 2 is lighter string

$f_{1}= \frac{1}{2l}\times\sqrt{\frac{T}{\mu_{1} }}$

and

$f_{2}= \frac{1}{2l}\times\sqrt{\frac{T}{\mu_{2} }}$

dividing f_1  by f_2 we get

now $\frac{f_{1}}{f_{2}} =\sqrt{\frac{\mu_2}{\mu_1} }$

= $\sqrt{\frac{\mu_2}{4\mu_2} }$

= $\frac{1}{2}$

⇒ $f_1=\frac{f_2}{2}$

hence the fundamental frequency of massive string is half of the fundamental frequency of lighter string