Question

The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lies below 1250Hz are:
(velocity of sound$= 340m{s^{ - 1}}$)
(A) 4
(B) 5
(C) 7
(D) 6

Answer 1

Hint We should know that oscillation is defined as the motion that is moving at a distance about the equilibrium position and repeats itself over and over for a period of time. Based on the concept the question is solved.

Complete step by step answer
The fundamental frequency of a closed organ pipe can be denoted as
$\upsilon = \dfrac{v}{{4L}}$
It is given that $v = 340m{s^{ - 1}}$.
$L = 85cm = 0.85m$
$\therefore \upsilon = \dfrac{{340m{s^{ - 1}}}}{{4 \times 0.85m}} = 100Hz$
The natural frequencies of a closed organ pipe can be listed from the given equation as:
${\upsilon _n} = \left( {2n - 1} \right)\upsilon$
$= \upsilon ,3\upsilon ,5\upsilon ,7\upsilon ,9\upsilon ,11\upsilon ,13\upsilon ,...$
$= 100Hz,300Hz,500Hz,700Hz,900Hz,1100Hz,1300Hz,...$ and so on.
Hence, we can see the natural frequencies that lie below the 1250Hz is 6.

Therefore, the correct answer is option D.

Note It is known that a motion repeating itself is referred to as the periodic or oscillatory motion. The object in the motion oscillates about an equilibrium position which is due to the restoring force or torque.