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  • When a vibrating tunning fork of frequency 512 Hz is hard above the mouth of a resonance tube of adjustable length, the first two successive position of resonance occur, when the length of the air

When a vibrating tunning fork of frequency 512 Hz is hard above the mouth of a resonance tube of adjustable length, the first two successive position of resonance occur, when the length of the air columns are 15.4 cm and 48.6 cm respectively . Then, the velocity of sound is -

Option: 1

\mathrm{512(48.6-15.4) \mathrm{cm} / \mathrm{s}}


Option: 2

\mathrm{1024(48.6-15.4) \mathrm{cm} / \mathrm{s}}


Option: 3

\mathrm{256(48.6-15.4) \mathrm{cm} / \mathrm{s}}


Option: 4

\mathrm{2 \times 512(48.6+15.4) \mathrm{cm} / \mathrm{s}}


Answers (1)

best_answer

Diffrence in lengths will be equal to one loop or \frac{\lambda}{2}

\therefore \quad \lambda=2(48.6-15.4) \mathrm{cm}

Now,

\begin{aligned} & v=f \lambda \\ & v=512 \times 2(48.6-15.4) \\ & v=1024(48.6-15.4) \mathrm{cm} / \mathrm{s} . \end{aligned}

Posted by

Sanket Gandhi

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