A wire of length 2L, is made by joining two wires A and B of 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 :

Waves PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q58 with Solution

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Correct answer: C

Given data

Total wire length $=2L$
It is made of two equal halves, so each part has length $L$ .
Wire $A$ has radius $r$
Wire $B$ has radius $2r$
Both are of the same material, so density and Young's modulus are same.
The joint is a node.

Since the whole wire is stretched together, the tension $T$ is same in both parts.

Wave speed in each part

For a stretched wire,

v = \sqrt{\frac{T}{\mu}}

where $\mu$ is linear mass density.

Because both are of same material,

\mu \propto \text{area} \propto r^2

So,

\mu_A \propto r^2, \qquad \mu_B \propto (2r)^2 = 4r^2

Hence,

\mu_B = 4\mu_A

Therefore,

v_A = \sqrt{\frac{T}{\mu_A}}, \qquad v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{T}{4\mu_A}} = \frac{v_A}{2}

So,

v_A : v_B = 2:1

Condition of standing waves in each part

The joint is a node, and the ends of the full wire are also nodes for the normal mode of vibration. Thus each segment of length $L$ has nodes at both ends.

For a string fixed at both ends,

L = n\frac{\lambda}{2}

where $n$ is the number of antinodes.

So for wire $A$ ,

L = p\frac{\lambda_A}{2} quad\Rightarrow\quad \lambda_A = \frac{2L}{p}

For wire $B$ ,

L = q\frac{\lambda_B}{2} quad\Rightarrow\quad \lambda_B = \frac{2L}{q}

Same frequency for both parts

Since the whole wire vibrates together, frequency is same in both parts:

f = \frac{v_A}{\lambda_A} = \frac{v_B}{\lambda_B}

Substitute $\lambda_A$ and $\lambda_B$ :

\frac{v_A}{2L/p} = \frac{v_B}{2L/q}

\frac{p v_A}{2L} = \frac{q v_B}{2L}

p v_A = q v_B

Thus,

\frac{p}{q} = \frac{v_B}{v_A} = \frac{1}{2}

Hence,

p:q = 1:2

Check options

A: $3:5$ ❌
B: $4:9$ ❌
C: $1:2$ ✅
D: $1:4$ ❌

Final answer

\boxed{p:q = 1:2}

So the correct option is C.

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View written solutionFree

Correct answer: C

Given data

Total wire length $=2L$
It is made of two equal halves, so each part has length $L$ .
Wire $A$ has radius $r$
Wire $B$ has radius $2r$
Both are of the same material, so density and Young's modulus are same.
The joint is a node.

Since the whole wire is stretched together, the tension $T$ is same in both parts.

Wave speed in each part

For a stretched wire,

v = \sqrt{\frac{T}{\mu}}

where $\mu$ is linear mass density.

Because both are of same material,

\mu \propto \text{area} \propto r^2

So,

\mu_A \propto r^2, \qquad \mu_B \propto (2r)^2 = 4r^2

Hence,

\mu_B = 4\mu_A

Therefore,

v_A = \sqrt{\frac{T}{\mu_A}}, \qquad v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{T}{4\mu_A}} = \frac{v_A}{2}

So,

v_A : v_B = 2:1

Condition of standing waves in each part

The joint is a node, and the ends of the full wire are also nodes for the normal mode of vibration. Thus each segment of length $L$ has nodes at both ends.

For a string fixed at both ends,

L = n\frac{\lambda}{2}

where $n$ is the number of antinodes.

So for wire $A$ ,

L = p\frac{\lambda_A}{2} quad\Rightarrow\quad \lambda_A = \frac{2L}{p}

For wire $B$ ,

L = q\frac{\lambda_B}{2} quad\Rightarrow\quad \lambda_B = \frac{2L}{q}

Same frequency for both parts

Since the whole wire vibrates together, frequency is same in both parts:

f = \frac{v_A}{\lambda_A} = \frac{v_B}{\lambda_B}

Substitute $\lambda_A$ and $\lambda_B$ :

\frac{v_A}{2L/p} = \frac{v_B}{2L/q}

\frac{p v_A}{2L} = \frac{q v_B}{2L}

p v_A = q v_B

Thus,

\frac{p}{q} = \frac{v_B}{v_A} = \frac{1}{2}

Hence,

p:q = 1:2

Check options

A: $3:5$ ❌
B: $4:9$ ❌
C: $1:2$ ✅
D: $1:4$ ❌

Final answer

\boxed{p:q = 1:2}

So the correct option is C.

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