Question #117372
A student stands at a distance of 400 m from a wall and claps two pieces of wood. After the first clap the student claps whenever an echo is heard from the wall. Another student starts a stopwatch at the first clap and stops it after the twentieth clap. The stopwatch records a time of 50 s. Find the speed of sound. An object vibrating with a frequency 850 Hz causes resonance in a tube of air when the shortest length of air column is 10.3 cm. Calculate the second resonant length, assuming the speed of sound in air is 340 m/s.
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Expert's answer
2020-05-24T18:02:37-0400

If the speed of sound is v, the distance from the wall is d, the time between two claps will be


t=2dv.t=\frac{2d}{v}.

The student clapped 20 times, which means that the stopwatch stopped after 19 complete claps. The total time of the experiment was


T=50 s=19t=38dv, v=2dT=3840050=304 m/s.T=50\text{ s}=19t=\frac{38d}{v},\\ \space\\ v=\frac{2d}{T}=\frac{38\cdot400}{50}=304\text{ m/s}.

The value is quire low, the experiment was conducted either in cold weather or high in the mountains.

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A resonance is a rapid increase of amplitude when the value of an external frequency approaches to the value of an intrinsic frequency. For one-end-closed pipe the first resonant frequency is


f1=v4L=34040.103=825 Hz.f_1=\frac{v}{4L}=\frac{340}{4\cdot0.103}=825\text{ Hz}.

The resonant length for a tube with both ends opened will be


Ls=v4f1=3404850=0.2 m.L_s=\frac{v}{4f_1}=\frac{340}{4\cdot850}=0.2\text{ m}.


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