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An air column in a tube $32 \mathrm{~cm}$ long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length $66 \mathrm{~cm}$ 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)
Options:
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)
Solution:
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Verified Answer
The correct answer is:
$264 \mathrm{~Hz}, 256 \mathrm{~Hz}$
We knows frequency of a closed end an column
$$
n_1=\frac{v}{4 / 1}
$$
We knows frequency of a open end an column
$$
n_2=\frac{v}{2 l_2}
$$
Given, $l_1=32 \mathrm{~cm}, I_2=66 \mathrm{~cm}$
and $\quad n_1-n_2=8$ heat $/ \mathrm{s}$
So, $\quad n_1=\frac{v}{4 \times 32}=\frac{v}{128}$
and $\quad n_2=\frac{v}{2 \times 66}=\frac{v}{132}$
In given condition,
$$
\begin{aligned}
& \frac{v}{128}-\frac{v}{132}=8 \\
& v=8448 \times 4 \\
& v=33792 \\
& \text { Hence, } \quad n_1=\frac{33792}{128} \\
& n_1=264 \mathrm{~Hz} \\
& \text { and } \\
& n_2=\frac{33792}{132} \\
& n_2=256 \mathrm{~Hz} \\
&
\end{aligned}
$$
$$
n_1=\frac{v}{4 / 1}
$$
We knows frequency of a open end an column
$$
n_2=\frac{v}{2 l_2}
$$
Given, $l_1=32 \mathrm{~cm}, I_2=66 \mathrm{~cm}$
and $\quad n_1-n_2=8$ heat $/ \mathrm{s}$
So, $\quad n_1=\frac{v}{4 \times 32}=\frac{v}{128}$
and $\quad n_2=\frac{v}{2 \times 66}=\frac{v}{132}$
In given condition,
$$
\begin{aligned}
& \frac{v}{128}-\frac{v}{132}=8 \\
& v=8448 \times 4 \\
& v=33792 \\
& \text { Hence, } \quad n_1=\frac{33792}{128} \\
& n_1=264 \mathrm{~Hz} \\
& \text { and } \\
& n_2=\frac{33792}{132} \\
& n_2=256 \mathrm{~Hz} \\
&
\end{aligned}
$$
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