Question

12 195 Hz
If length, tension, density and radius of cross section
of a stretched string A are four times of that of
stretched string B, then ratio of fundamental
frequency of A and B is
(1) 1:2
(2) 1 : 16 | u
(3) 8:1
4
:4​

Answer 1

Answer:

Ratio of fundamental frequency in two wires is given as

$\frac{f_A}{f_B} = 1 : 16$

Explanation:

As we know that the fundamental frequency in the stretched string is given as

$f_o = \frac{1}{2L}\sqrt{\frac{T}{\rho \pi r^2}}$

now we know that

Length, Density, Tension and Radius of wire A is 4 times that of wire B

so we will have

$\frac{f_A}{f_B} = \frac{L_B}{L_A}\sqrt{\frac{T_A}{T_B} \frac{\rho_B}{\rho_A} \frac{r_B^2}{r_A^2}}$

now we have

$\frac{f_A}{f_B} = \frac{1}{4}\sqrt{\frac{4}{1} \frac{1}{4} \frac{1^2}{4^2}}$

$\frac{f_A}{f_B} = 1 : 16$

#Learn

Topic : Fundamental frequency in string

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