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  1. Problems
  2. /The speed of sound in air is 331 m/s at 0 °C, and it grows with the square root of the kelvin…
MediumPhysicsHigh School (11–12)Waves & Sound › Sound Waves and HearingWaves & Sound › Superposition and Standing WavesNumericalpythonNEETAEEE (Amrita)AMUEEEAP EAPCETBCECECOMEDK UGET
+14 more exams
GUJCETJCECEJKCETKCETKEAMKIITEELPUNESTMET (Manipal)MHT-CETOJEESRMJEEETS EAPCETUPESEATVITEEE

The speed of sound in air is 331 m/s at 0 °C, and it grows with the square root of the kelvin…

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.

The speed of sound in air is 331 m/s331\ \text{m/s}331 m/s at 0 ∘C0\ ^\circ\text{C}0 ∘C, and it grows with the square root of the kelvin temperature. A fork of 480 Hz480\ \text{Hz}480 Hz is used on a tube with an end correction of 1.0 cm1.0\ \text{cm}1.0 cm. Find the length of the first resonance (a) on a day at 17 ∘C17\ ^\circ\text{C}17 ∘C and (b) on a day at 37 ∘C37\ ^\circ\text{C}37 ∘C.

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cm
cm
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) At 17 ∘C17\ ^\circ\text{C}17 ∘C. T=290 KT = 290\ \text{K}T=290 K, so v=331 m/s×290 K273 K=341.2 m/sv = 331\ \text{m/s} \times \sqrt{\dfrac{290\ \text{K}}{273\ \text{K}}} = 341.2\ \text{m/s}v=331 m/s×273 K290 K​​=341.2 m/s.

l1=341.2 m/s4×480 Hz−0.010 m=0.1777 m−0.010 m=16.8 cm\displaystyle l_1 = \frac{341.2\ \text{m/s}}{4 \times 480\ \text{Hz}} - 0.010\ \text{m} = 0.1777\ \text{m} - 0.010\ \text{m} = 16.8\ \text{cm}l1​=4×480 Hz341.2 m/s​−0.010 m=0.1777 m−0.010 m=16.8 cm

(b) At 37 ∘C37\ ^\circ\text{C}37 ∘C. T=310 KT = 310\ \text{K}T=310 K, so v=352.7 m/sv = 352.7\ \text{m/s}v=352.7 m/s.

l1=352.7 m/s4×480 Hz−0.010 m=0.1837 m−0.010 m=17.4 cm\displaystyle l_1 = \frac{352.7\ \text{m/s}}{4 \times 480\ \text{Hz}} - 0.010\ \text{m} = 0.1837\ \text{m} - 0.010\ \text{m} = 17.4\ \text{cm}l1​=4×480 Hz352.7 m/s​−0.010 m=0.1837 m−0.010 m=17.4 cm

Check. On the hotter day sound is faster, the wavelength is longer, and the water must be about 6 mm6\ \text{mm}6 mm lower.

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