A particle smaller than an atom or a <span>cluster of such particles </span>
Answer: was it this problem?
Explanation:
Length of the pipe = 0.39 m
Third harmonic frequency = 1400 Hz
For the third harmonic:
Wavelength = 
The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength
Distance from center = 0.25 × wavelength
Distance = 
Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm
Distance = 
Distance from the center to the nearest anti - node = 6.5 cm
Hence, the nearest distance to the anti - node from the center = 6.5 cm
So, option C is correct.
V=wave velocity , <span>f= frequency, </span><span>λ=wavelength </span>
<span>Use it to find corresponding wavelengths for</span><span> f=28 Hz </span>
<span>λ= v/f= 337/28=12.036 m
</span>
<span>for f=4200 Hz </span>
<span>λ= v/f=337/4200= 0.08 m </span>
<span>So max. wavelength is 12.036 m and </span>
<span>Min Wavelength is 0.08 m </span>
<span>So the range is between .08 m and 12.036 m
</span>Hope this helps.