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)

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

Tardigrade Admin · Accepted Answer

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 c m, l_{2} = 66 c m

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