JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Waves & Sound

A wire having a linear mass density

9.0 \times 1 0^{- 4} kg/m

is stretched between two rigid supports with a tension of 900 N. The wire resonates at a frequency of 500 Hz. The next higher frequency at which the same wire resonates is 550 Hz. The length of the wire is _____ m.

Answer: 10

Solution

The wave speed on the wire is

v = \frac{T}{μ} = \frac{900}{9.0 \times 1 0 ^{- 4}} = 1000 m s^{- 1}

. Consecutive resonances of a wire fixed at both ends differ by the fundamental

\frac{v}{2 L}

, so

550 - 500 = \frac{v}{2 L}

. Hence

L = \frac{v}{2 ( 50 )} = \frac{1000}{100} = 10 m .

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Integer answer

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