Question

An air column in a tube $32\, cm$ long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length $66\, cm$ is in resonance with another tuning fork. When these two tuning forks are sounded together, they produce 8 beats per second. Then the frequencies of the two tuning forks are, (Consider fundamental frequencies only)

• A. 250 Hz , 258 Hz
• B. 240 Hz , 248 Hz
• C. 264 Hz , 256 Hz
• D. 280 Hz , 272 Hz

Answer 1

Answer: (C)

We knows frequency of a closed end an column
$n_{1}=\frac{V}{4 l_{1}}$
We knows frequency of a open end an column
$n_{2}=\frac{V}{2 l_{2}}$
Given, $l_{1}=32\, cm ,\, l_{2}=66\, cm$
and $n_{1}-n_{2}=8$ heat $/ s$
So, $n_{1}=\frac{v}{4 \times 32}=\frac{V}{128}$
and $n_{2}=\frac{v}{2 \times 66}=\frac{v}{132}$
In given condition,
$\frac{v}{128}-\frac{v}{132}=8$
$v=8448 \times 4$
$v =33792$
Hence, $n_{1} =\frac{33792}{128}$
$n_{1} =264\, Hz$
and $n_{2} =\frac{33792}{132}$
$n_{2} =256\, Hz$