Question

The number of possible natural oscillations of air
column in a pipe closed at one end of length 85 cm
whose frequencies below 1250 Hz w (velocity
of sound 340 ms-1​

Answer 1

Answer:

The possible natural frequencies of the pipe is 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz

Explanation:

As we know that the frequency of oscillation in the closed end pipe is given as

$f = \frac{(2N + 1)v}{4L}$

speed of the sound wave in the air is given as

$v = 340 m/s$

$L = 85 cm$

Now we have

$f = \frac{(2N + 1)(340)}{4(0.85)}$

$f = (2N + 1)100$

$1250 = (2N + 1)100$

so N is less than or equal to 5

so we have

N = 0

$f_1 = 100 Hz$

N = 1

$f_2 = 300 Hz$

N = 2

$f_3 = 500 Hz$

N = 3

$f_4 = 700 Hz$

N = 4

$f_5 = 900 Hz$

N = 5

$f_6 = 1100 Hz$

#Learn

Topic : Closed organ pipe

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