Question 124: Waves

Question

In an open organ pipe $
u_3$ and $
u_6$ are $3^{	ext {rd }}$ and $6^{	ext {th }}$ harmonic frequencies, respectively. If $
u_6-
u_3=2200 \mathrm{~Hz}$ then length of the pipe is $\_\_\_\_$ mm .
(Take velocity of sound in air is $330 \mathrm{~m} / \mathrm{s}$.)

Accepted Answer

For an open organ pipe (open at both ends), the standing waves formed have antinodes at both ends.
The frequency of the $\mathrm{n}^{	ext {th }}$ harmonic $\left(
u_{\mathrm{n}}ight)$ is an integer multiple of the fundamental frequency and is given by the formula :
$$ \mathrm{v}_{\mathrm{n}}=\frac{\mathrm{nv}}{2 \mathrm{~L}} $$
Where :
n is the harmonic number (integer: $1,2,3 \ldots$ )
$v$ is the velocity of sound in the medium
L is the length of the open pipe
Using the formula for the $\mathrm{n}^{	ext {th }}$ harmonic :
Frequency of the $3^{	ext {rd }}$ harmonic $=v_3=\frac{3 v}{2 L}$
Frequency of the $6^{	ext {th }}$ harmonic $=v_6=\frac{6 \mathrm{v}}{2 \mathrm{~L}}$
The difference between the frequencies of $6^{	ext {th }}$ and $3^{	ext {rd }}$ harmonic is 2200 Hz ,
$$ v_6-v_3=2200 $$
$\Rightarrow \frac{6 \mathrm{v}}{2 \mathrm{~L}}-\frac{3 \mathrm{v}}{2 \mathrm{~L}}=2200$
$\Rightarrow \frac{3 \mathrm{v}}{2 \mathrm{~L}}=2200$
$\Rightarrow \frac{3 	imes 330}{2 L}=2200$
$\Rightarrow \frac{990}{2 \mathrm{~L}}=2200$
$$ \Rightarrow L=\frac{990}{2200 	imes 2} \Rightarrow L=0.225 \mathrm{~m}=225 \mathrm{~mm} $$
Therefore, the length of the pipe is 225 mm. Hence, the correct option is (B).

Answer

200

Answer

225

Answer

275

Answer

250

Waves

In an open organ pipe \nu_3 and \nu_6 are 3^{\text {rd }} and 6^{\text {th }} harmonic frequencies, respectively. If \nu_6-\nu_3=2200 \mathrm{~Hz} then length of the pipe is \_\_\_\_ mm . (Take velocity of sound in air is 330 \mathrm{~m} / \mathrm{s}.)

Choose the correct answer: