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  1. Problems
  2. /In a resonance tube experiment a fork of 512 Hz gives the first resonance at 15.9 cm and the second…
EasyPhysicsHigh School (11–12)Waves & Sound › Sound Waves and HearingWaves & Sound › Superposition and Standing WavesNumerical

In a resonance tube experiment a fork of 512 Hz gives the first resonance at 15.9 cm and the second…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

In a resonance tube experiment a fork of 512 Hz512\ \text{Hz}512 Hz gives the first resonance at 15.9 cm15.9\ \text{cm}15.9 cm and the second at 49.5 cm49.5\ \text{cm}49.5 cm. Find (a) the wavelength, (b) the speed of sound, and (c) the end correction.

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cm
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Need a reminder? The key ideaShowHide

A resonance tube (closed by water) booms when l1+e=λ/4l_1 + e = \lambda/4l1​+e=λ/4, l2+e=3λ/4l_2 + e = 3\lambda/4l2​+e=3λ/4, and so on. So

v=fλ=2f(l2−l1)e=l2−3l12≈0.3d\displaystyle v = f\lambda = 2f(l_2 - l_1) \qquad e = \frac{l_2 - 3l_1}{2} \approx 0.3dv=fλ=2f(l2​−l1​)e=2l2​−3l1​​≈0.3d

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

(a) λ=2(49.5 cm−15.9 cm)=67.2 cm\lambda = 2(49.5\ \text{cm} - 15.9\ \text{cm}) = 67.2\ \text{cm}λ=2(49.5 cm−15.9 cm)=67.2 cm.

(b) v=fλ=512 Hz×0.672 m=344 m/sv = f\lambda = 512\ \text{Hz} \times 0.672\ \text{m} = 344\ \text{m/s}v=fλ=512 Hz×0.672 m=344 m/s.

(c) e=49.5 cm−3×15.9 cm2=49.5 cm−47.7 cm2=0.9 cme = \dfrac{49.5\ \text{cm} - 3 \times 15.9\ \text{cm}}{2} = \dfrac{49.5\ \text{cm} - 47.7\ \text{cm}}{2} = 0.9\ \text{cm}e=249.5 cm−3×15.9 cm​=249.5 cm−47.7 cm​=0.9 cm.

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