Question

An open organ pipe vibrating in second
harmonic having frequency fr. Its one end is
closed & Grequency slowly increases to f7 &
vibrates in nth harmonic then possible values of
n & -
4
(A)n=3; f = (B)n=3
(C) n=5; f = 9 (D) nas:​

Answer 1

Explanation:

An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f1. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from f1 then again a resonance is obtained with a frequency f2. If in this case the pipe vibrates nth harmonics then

Answer

n = 5, f2 = (5/4)f1

Answer 2

Question :--- An open organ pipe vibrating in second harmonic having frequency fr. Its one end is closed & Grequency slowly increases to f7 &

vibrates in nth harmonic then possible values of n is ?

Concept :--

→ For an open organ pipe, the length L is given as

L = nλ/2

where, λ is the wavelength of wave and n is an integer and by putting n=1,2,3..............we get the modes of vibration.

n=1 gives first harmonics, n=2 gives second harmonics and so on.

Here, an open organ pipe of length L vibrates in second harmonic mode ...

__________________________

Solution :---

Fundamental harmonic is one of the organ pipe is given by :---

→ L = (λ1 / 2) and V1 = (V / λ1) = (V / 2L)

Now the tube vibrates in second harmonic, is :--

→ f1 = 2V1 = (2V / 2L) = (V/L)

if one end is close,the old odd harmonics & Fundamental frequency

= (V / 4L)

And the other harmonics are (3V / 4L), (5V / 4L).

_____________________

one of the frequency starts increasing the first higher harmonic that is resonated to (3V / 4L)

if n = 3

→ f2 = (3V / 4L) = (3/4)f1

The frequency is increased from (V/L) hence (3/4).

f2 is not greater than f1 [ as f1 = (V/L)].

Therefore f1 (5/4) is the correct answer.

Where n = 5

and

f2 = 5 × (V / 4L) = (5/4)f1