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A tube closed at one end has a vibrating diaphragm at the other end, which may be assumed to be displacement node. It is found that when the frequency of the diaphragm is 2000 Hz, then a stationary wave pattern is set up in which the distance between adjacent nodes is 8 cm. When the frequency is gradually reduced, then the stationary wave pattern disappears but another stationary wave pattern reappears at a frequency of 1600 Hz. Calculate 

(i) The speed of sound in air,

(ii) The distance between adjacent nodes at a frequency of 1600 Hz,

(iii) The distance between the diaphragm and the closed end, and 

(iv) The next lower frequencies at which stationary wave pattern will be obtained.

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The standing waves generated inside the tube closed at one end, have the wavelength nλ = 2L  where L is length of the tube. The velocity of the wave in air is given by v = fλ, where n is the frequency of the sound wave. Since the node-to-node distance is

Since the node-to-node distance is

∴Distance between nodes = 0.2/2 = 0.1 m = 10 cm

(iii) Since there are nodes at the ends, the distance between the closed end and the diaphragm must be an 

integral multiple of λ/2

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