Question

In a resonance tube experiment, tuning fork A is in resonance with air column of length $40\text{cm}$ and tuning fork B is in resonance with air column of length $41\text{cm}$.
Two tuning forks A and B sounded together produce $10$ beats per second. Calculate the frequencies of forks.
(Assume fundamental mode for air column)

• A. $400\text{Hz},390\text{Hz}$
• B. $430\text{Hz},410\text{Hz}$
• C. $410\text{Hz},400\text{Hz}$
• D. $410\text{Hz},420\text{Hz}$

Answer 1

The correct option is C $410\text{Hz},400\text{Hz}$

Let the frequency of the first fork be $f_1$ and that of second be $f_2$.
For tube closed at one end,
$f_1=\frac{\upsilon }{4L_1}=\frac{\upsilon }{4\times 40\times {10}^{-2}}$
and
$f_2=\frac{\upsilon }{4L_2}=\frac{\upsilon }{4\times 41\times {10}^{-2}}$
we can see
$f_1>f_2$
$\therefore f_1-f_2=10$ ...(i)
and
$\frac{f_1}{f_2}=\frac{4\times 41\times {10}^{-2}}{4\times 40\times {10}^{-2}}=\frac{41}{40}$ ...(ii)
putting the value of $f_2$ in eq (i)
So,
$f_1-\frac{40}{41}{f}_1=10$
$\Rightarrow \frac{f_1}{41}=10\Rightarrow f_1=410\text{Hz}$
$\therefore f_2=410-10=400\text{Hz}$