Question

A wire of length 2L is made by joining two wires A and B of the same length but different radii r and 2r, and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is ______.

Options

• 3 : 5

• 4 : 9

• 1 : 2

• 1 : 4

Answer 1

A wire of length 2L is made by joining two wires A and B of the same length but different radii r and 2r, and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is 1 : 2.

Explanation:

Given: The wire has a total length of 2L, where L is the length of the wire's A and B parts, r is the wire's A part's radius and 2r is the wire's B part, p is the number of antinodes in part A, and q is the number of antinodes in part B. The intersection of two wires is referred to as a node.

To find: p : q

Part A's mass per unit of length:

`mu_A = (rhopir^2L)/L` ....(i)

(ρ is density of wire A = density of wire B)

Part B mass per unit length:

`mu_B = (rhopi(2r)^2L)/L = 4mu_A` .......(ii)

Wave speed in Part A:

`v_A = sqrt(T/(mu_A))` ........(iii)

Wave speed in Part B:

`v_B = sqrt(T/mu_B) = 1/2sqrt(T/mu_A)` .........(iv)

Because the frequency of the wave in wire A is equal to the frequency of the wave in wire B.

`f_A = (pv_A)/(2L) = p/(2L)sqrt(T/mu_A)` ......(v)

`f_B = (pv_B)/(2L) = q/(4L)sqrt(T/mu_A)` ......(vi)

Equation solving (v) and (vi),

`p/(2L)sqrt(T/mu_A) = q/(4L)sqrt(T/mu_A)`

p : q = 1 : 2