Question 278: Waves

Question

Two pipes of lengths $\mathrm{L}_1$ and $\mathrm{L}_2$, open at both ends are joined in series. If ' $f_1$ ' and ' $f_2$ ' are the fundamental frequencies of two pipes, then the fundamental frequency of series combination will be (neglect end correction)

Accepted Answer

Let us solve it step by step using NCERT textbook concepts.
Step 1: Fundamental Frequency of Each Pipe
For an open pipe of length $L$, the fundamental frequency is given by:
$$ f = \frac{v}{2L} $$
where $v$ is the speed of sound in air.
So, for two pipes:
For first pipe of length $L_1$, $f_1 = \dfrac{v}{2L_1}$
For second pipe of length $L_2$, $f_2 = \dfrac{v}{2L_2}$
Step 2: Combined Length
When the two pipes are joined in series, the effective length is the sum of both:
$$ L = L_1 + L_2 $$
Step 3: Fundamental Frequency of Combined Pipe
For the combined pipe (length $L$), the new fundamental frequency, $f$, will be:
$$ f = \frac{v}{2L} = \frac{v}{2(L_1 + L_2)} $$
Step 4: Relate $L_1$ and $L_2$ to $f_1$ and $f_2$
From the above,
$ L_1 = \frac{v}{2f_1} \ L_2 = \frac{v}{2f_2} $
Therefore,
$ L_1 + L_2 = \frac{v}{2f_1} + \frac{v}{2f_2} = \frac{v}{2}\left(\frac{1}{f_1} + \frac{1}{f_2}ight) $
Step 5: Substitute Back into $f$
So,
$ f = \frac{v}{2(L_1 + L_2)} = \frac{v}{2 	imes \frac{v}{2} \left(\frac{1}{f_1} + \frac{1}{f_2}ight)} $
The $v/2$ in numerator and denominator cancel:
$ f = \frac{1}{\left(\frac{1}{f_1} + \frac{1}{f_2}ight)} $
$ f = \frac{1}{\left(\frac{f_2+f_1}{f_1 f_2}ight)} = \frac{f_1 f_2}{f_1 + f_2} $
Step 6: Check the Options
Option C: $\boxed{\dfrac{f_1 f_2}{f_1 + f_2}}$
Correct Answer: Option C

Answer

$\frac{f_1 f_2}{f_1-f_2}$

Answer

$\mathrm{f}_1+\mathrm{f}_2$

Answer

$\frac{f_1 f_2}{f_1+f_2}$

Answer

$\sqrt{\mathrm{f}_1^2+\mathrm{f}_2^2}$

Waves

Two pipes of lengths \mathrm{L}_1 and \mathrm{L}_2, open at both ends are joined in series. If ' f_1 ' and ' f_2 ' are the fundamental frequencies of two pipes, then the fundamental frequency of series combination will be (neglect end correction)

Choose the correct answer: