fundamental frequency
L=4λv1=4Lv
L=2λ+4λ=43λ
v2=4L3v
∴v=odd(4Lv)
Now L=85cm=0.85 m
v=340m/s
=odd(4×0.85340)
=odd×100Hz
Hence possible frequencies below 1250 Hz are
100 Hz, 300 Hz, 500 Hz...........1100 Hz
Hence 6
A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
Held on 6 Apr 2014 · Verified 6 Jul 2026.
12
8
6
4
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