1. Tardigrade
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  3. Physics
  4. Two tuning forks A and B vibrating simultaneously produce 5 beats. Frequency of B is 512 Hz. It is seen that if one arm of A is filed, then the number of beats increases. The frequency of A will be

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Q. Two tuning forks $A$ and $B$ vibrating simultaneously produce $5$ beats. Frequency of $B$ is $512 \,Hz$. It is seen that if one arm of $A$ is filed, then the number of beats increases. The frequency of $A$ will be

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Solution:

There are five beats between $A$ and $B$, therefore, the possible frequencies of $A$ are
$512 \pm 5=(517$ or $507) Hz$
When one prong of $A$ is filed its frequency becomes greater than the original frequency. If we assume that the original frequency of $A$ is $517\, Hz$, then on filing its frequency will be greater than $517\, Hz$. The beats between $A$ and $B$ will be more than $5 .$ But it is given that the beats are increasing so it is only possible if frequency of $A$ is $517 \,Hz$