Question

Two stretched wires A and B of the same lengths vibrate independently. If the radius, density and
tension of wire A are respectively twice those of wire B, then the fundamental frequency of vibration o
A relative to that of B is
[REE - 1990)
(A) 1:1
(B) 1:2
(C) 1:4
(D) 1:8
​

Answer 1

Answer:

the fundamental frequency of vibration o f wire A relative to that of B is 1 : 2

Explanation:

As we know that the fundamental frequency in the wire is given as

$f_o = \frac{1}{2L}\sqrt{\frac{T}{\rho A}}$

now we know that radius, density and tension of wire A is double that of wire B

Also it is given that the length of two wires is same

so the ratio of fundamental frequency of two wires is given as

$\frac{f_a}{f_b} = \frac{\frac{1}{2L}\sqrt{\frac{T_a}{\rho_a A_a}}}{\frac{1}{2L}\sqrt{\frac{T_b}{\rho_b A_b}}}$

now we have

$\frac{f_a}{f_b} = \sqrt{(\frac{T_a}{T_b})(\frac{\rho_b}{\rho_a})(\frac{A_b}{A_a})}$

$\frac{f_a}{f_b} = \sqrt{2 \times \frac{1}{2} \times \frac{1}{4}}$

$\frac{f_a}{f_b} = \frac{1}{2}$

#Learn

Topic : Standing waves in a wire

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