Question

A wire of length 2L is made by joining two wires A and b of same lengths 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:
(A) 1 : 4 (B) 1 : 2
(C) 3 : 5 (D) 4 : 9

Answer 1

Let mass per unit length of wires are μ1 and μ2 respectively.

∵ Materials are same, so density ρ is same.

∴μ1=Lρπr2L=μ and μ2=Lρ4πr2L=4μ

Tension in both are same =T, let speed of wave in wires are V1 and V2

V1=μT=V.V2=4μT=2V

So fundamental frequencies in both wires are f01=2LV1=2LV and f02=2LV2

Answer 2

Ratio of no. of antinodes is =  1 : 2

Step by step explanation:

It is given that two wires of same length L are joined to make a wire of length 2L.

If the material of both the wires is same then their mass density will also be same = ρ

As we know that

frequency,  f = $\frac{velocity}{length}$                                                       (equation 1)

Velocity,     v = $\sqrt{\frac{Tension}{\text{mass per unit length}}}$                                    (equation 2)

mass per unit length of rod A = $\frac{\rho \pi (r)^2L}{L} = \rho \pi (r)^2$  = M     (equation 3)

mass per unit length of rod B =$\frac{\rho \pi (2r)^2L}{L} =4 \rho \pi (r)^2$ = 4M   (equation 4)

Both wires works as a big wire of doubl length, It has same tension in both wire which is equal to T

So,( Putting equation 3 in equation 2, we get)

Velocity of rod A = $\sqrt{\frac{T}{\text{M}}}$ =   v                                           (equation 5)

( Putting equation 4 in equation 2, we get)

Velocity of rod B = $\sqrt{\frac{T}{\text{4M}}} = \frac{v}{2}$                                             (equation 6)

Now, (Putting value of equation 5 in equation 1)

The frequency of rod A = $\textbf{\large f}_0_1 = \frac{v}{2L}$   = f

(Putting value of equation 6 in equation 1)

The frequency of rod B = $\textbf{\large f}_0_2 = \frac{(\frac{v}{2})}{2L} = \frac{f}{2}$

Frequency at which both resonate is L.C.M of both frequencies like $\frac{v}{2L}$

​

So, the no. of loops in wires are 1 and 2 .

So, $\textbf{\Large Ratio of no. of antinodes is = 1:2}$