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Standard VIII

Physics

Compression and Rarefaction

A tuning fork...

Question

A tuning fork A of frequency $510 Hz$ produces $4$ beats per second when sounded with another tuning fork B of unknown frequency. If B is loaded with wax, the number of beats is again $4$ per second. The frequency of the tuning fork B before it was loaded is

510 Hz

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514 Hz

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512 Hz

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506 Hz

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Solution

The correct option is B $514 Hz$
$f_{B} - f_{A} = 4$
After loading $f_{s}$ decreases and now $f_{A} - f_{B} = 4$
Hence, $f_{B} = 4 + 510 = 514 Hz$ (before loading)

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A tuning fork A of frequency 510 Hz produces 4 beats per second when sounded with another tuning fork B of unknown frequency. If B is loaded with wax, the number of beats is again 4 per second. The frequency of the tuning fork B before it was loaded is

The correct option is B 514 HzfB−fA=4 After loading fs decreases and now fA−fB=4 Hence, fB=4+510=514 Hz (before loading)

A tuning fork A of frequency $510 Hz$ produces $4$ beats per second when sounded with another tuning fork B of unknown frequency. If B is loaded with wax, the number of beats is again $4$ per second. The frequency of the tuning fork B before it was loaded is

The correct option is B $514 Hz$
$f_{B} - f_{A} = 4$
After loading $f_{s}$ decreases and now $f_{A} - f_{B} = 4$
Hence, $f_{B} = 4 + 510 = 514 Hz$ (before loading)