JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Waves & Sound

A wire of length 2 L is made by joining two wires A and b of same lengths but different radii

r

and

2 r

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

p

and that in

B

q

then the ratio

p : q

is:

Answer: (B)

Solution

Both wires are of the same material and length

L

and carry the same tension, so they differ only in linear mass density, which scales as the square of the radius: wire

B

has twice the radius of

A

, so

μ_{B} = 4 μ_{A}

. The wave speed

v = T / μ

then makes

v_{B} = \frac{1}{2} v_{A}

. With a node forced at the joint, each wire holds a whole number of loops at the common frequency

f = \frac{p v _{A}}{2 L} = \frac{q v _{B}}{2 L}

. Cancelling gives

p v_{A} = q v_{B} = q \frac{v _{A}}{2}

, so

p : q = 1 : 2

. The antinode count equals the loop count, so the ratio of antinodes is

1 : 2

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option