An open pipe is in resonance in its 2 text nd 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 n text th harmonics then

Question

An open pipe is in resonance in its 2 text nd 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 n text th harmonics then (A) n=3, f2=(3/4) f1. (B) n=3, f2=(5/4) f1 (C) n=5, f2=(5/4) f1 (D) n=5, f2=(3/4) f1

Tardigrade Admin · Accepted Answer

Open pipe resonance frequency f1=(2 v/2 L) Closed pipe resonance frequency f2=(n v/4 L) f2=(n/4) f1 (where n is odd and f2>f1) ∴ n=5

$n = 3, f_{2} = \frac{3}{4} f_{1}$ .

$n = 3, f_{2} = \frac{5}{4} f_{1}$

$n = 5, f_{2} = \frac{5}{4} f_{1}$

$n = 5, f_{2} = \frac{3}{4} f_{1}$

Open pipe resonance frequency $f_{1} = \frac{2 v}{2 L}$
Closed pipe resonance frequency $f_{2} = \frac{n v}{4 L}$
$f_{2} = \frac{n}{4} f_{1}$ (where $n$ is odd and $f_{2} > f_{1})$
$∴ n = 5$

Q. 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

n=3,f2​=43​f1​.

n=3,f2​=45​f1​

n=5,f2​=45​f1​

n=5,f2​=43​f1​

Open pipe resonance frequency f1​=2L2v​ Closed pipe resonance frequency f2​=4Lnv​ f2​=4n​f1​ (where n is odd and f2​>f1​) ∴n=5

$n = 3, f_{2} = \frac{3}{4} f_{1}$ .

$n = 3, f_{2} = \frac{5}{4} f_{1}$

$n = 5, f_{2} = \frac{5}{4} f_{1}$

$n = 5, f_{2} = \frac{3}{4} f_{1}$

Open pipe resonance frequency $f_{1} = \frac{2 v}{2 L}$
Closed pipe resonance frequency $f_{2} = \frac{n v}{4 L}$
$f_{2} = \frac{n}{4} f_{1}$ (where $n$ is odd and $f_{2} > f_{1})$
$∴ n = 5$