JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Waves & Sound

A wire of length L and mass per unit length

6.0 \times 1 0^{- 3} kg m^{- 1}

is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz . Then L in meters is

Answer: (C)

Solution

Consecutive resonant frequencies of a string fixed at both ends differ by the fundamental, so

f_{0} = 490 - 420 = 70 Hz

. The fundamental is

f_{0} = \frac{1}{2 L} \frac{T}{μ}

, and with

T = 540 N

and

μ = 6.0 \times 1 0^{- 3} kg/m

the wave speed is

\frac{540}{6.0 \times 1 0 ^{- 3}} = 90000 = 300 m/s

. Thus

70 = \frac{300}{2 L}

, which gives

L = \frac{300}{140} \approx 2.1 m

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option