JEE Main 2023 · Physics · Waves and Sound · easy

In a resonance-tube experiment with a tuning fork of frequency 400 Hz, the first and second resonances occur at air-column lengths of 20 cm and 62…

Numerical answer

In a resonance-tube experiment with a tuning fork of frequency

400\ \mathrm{Hz}

, the first and second resonances occur at air-column lengths of

20\ \mathrm{cm}

and

62\ \mathrm{cm}

. The speed of sound in air (in

\mathrm{m\,s^{-1}}

) obtained from this experiment is

Show answer and solution

Answer: 336

The tube is closed at one end, so successive resonances differ by half a wavelength:

\dfrac\lambda2=62-20=42\ \mathrm{cm}

, giving

\lambda=0.84\ \mathrm m

. The speed is

v=f\lambda=400\times0.84=336\ \mathrm{m\,s^{-1}}

. The end correction is

e=\dfrac{62-3(20)}{2}=1\ \mathrm{cm}

and need not be known, which is why the difference method is used (using

\lambda=4\times20\ \mathrm{cm}

would give

320

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