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

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40 cm

80 cm

120 cm

160 cm

answer is A.

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Detailed Solution

$In first case, V = nλ V = 2000 \times 16 = 320 m / s (∵ λ / 2 = 8) In second case, n = 1600 H z λ = \frac{320}{1600} = 0 \cdot 2 m = 20 cm Let in the first case, it is p^{th} harmonic. Then in second case it will be (p - 1)^{th} harmonic. If N is the fundamental, pN = 2000 and (p - 1) = 1600 N = 400 Now L = λ / 2 or λ = 2 L So 32000 = 400 \times 2 L or L = 40 cm$