When a tuning...

Question

When a tuning fork of frequency 341 Hz is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first fork, the number of beats is two per second. The natural frequency of the second tuning fork is

335 Hz

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338 Hz

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344 Hz

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347 Hz

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Solution

The correct option is D 347 Hz
$x = 6 b p s$ , which decreases (i.e, $x ↓$ ) after second fork is loaded with wax.

Hence,
$| n_{B} ↓ - n_{A} | = | x ↓ |$
$⟹ n_{b} = n_{A} + x = 341 + 6 = 347 H z .$

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Similar questions

Q. When a tuning fork of frequency 341 Hz is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first fork, the number of beats is two per second. The natural frequency (in

Hz

)of the second tuning fork is

When a tuning fork of frequency 341 is sounded with another tuning fork, six

beats per second are heard. When the second tuning fork is loaded with wax

and sounded with the first tuning fork, the number of beats is two per

second. The natural frequency of the second tuning fork is

[MP PET 1989]

Q. A tuning fork

A

of unknown frequency produces

4

beats per second when sounded with another tuning fork

B

of frequency

254 Hz

. When tuning fork

A

is loaded with wax, it still gives

4

beats per second with tuning fork

B

. The initial unknown frequency of tuning fork

A

Q. A tuning fork

A

of unknown frequency produces

4

beats per second when sounded with another tuning fork

B

of frequency

254 Hz

. When tuning fork

A

is loaded with wax, it still gives

4

beats per second with tuning fork

A

. The initial unknown frequency of tuning fork

A

Q. A tuning fork

A

of frequency

512 H z

produces

5

beats per second when sounded with another tuning fork

B

of unknown frequency. If

B

is located with wax the number of beats is again

5

per second. The frequency of the tuning fork

B

before it was loaded is

When a tuning fork of frequency 341 Hz is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first fork, the number of beats is two per second. The natural frequency of the second tuning fork is

The correct option is D 347 Hzx=6 bps, which decreases (i.e, x↓) after second fork is loaded with wax. Hence, |nB↓−nA|=|x↓| ⟹nb=nA+x=341+6=347Hz.

The correct option is D 347 Hz
$x = 6 b p s$ , which decreases (i.e, $x ↓$ ) after second fork is loaded with wax.

Hence,
$| n_{B} ↓ - n_{A} | = | x ↓ |$
$⟹ n_{b} = n_{A} + x = 341 + 6 = 347 H z .$