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Standard XII

Physics

Periodic Motion

Two sitar str...

Question

Two sitar strings, $A$ and $B$ , playing the note 'Dha' are slightly out of tune and produce beats and frequency $5 H z$ . The tension of the string $B$ is slightly increased and the beat frequency is found to decrease by $3 H z$ . If the frequency of $A$ is $425 H z$ , the original frequency of $B$ is

430 H z

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428 H z

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422 H z

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420 H z

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Solution

The correct option is D $420 H z$
the difference in frequency is known as the number of beats.

Here,frequency of A $f_{A}$ = 425 Hz
We know,
Frequency of B $f_{B}$ = $f_{A}$ ± $b e a t f r e q u e n c y$
= 425 ± 5
= 420Hz or 430Hz
Frequncy decreases

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Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats and frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease by 3 Hz. If the frequency of A is 425 Hz, the original frequency of B is

The correct option is D 420 Hzthe difference in frequency is known as the number of beats.Here,frequency of A fA = 425 Hz We know, Frequency of B fB =fA ± beatfrequency= 425 ± 5 = 420Hz or 430Hz Frequncy decreases

The correct option is D $420 H z$
the difference in frequency is known as the number of beats.

Here,frequency of A $f_{A}$ = 425 Hz
We know,
Frequency of B $f_{B}$ = $f_{A}$ ± $b e a t f r e q u e n c y$
= 425 ± 5
= 420Hz or 430Hz
Frequncy decreases