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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a
40 cm
b
80 cm
c
120 cm
d
160 cm
answer is A.
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Detailed Solution
In first case, V=nλ V=2000×16=320m/s (∵λ/2=8) In second case, n = 1600 Hz λ=3201600=0⋅2m=20cm Let in the first case, it is pth 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=2L So 32000=400×2L or L=40cm