Question

Two tuning forks A and B are in unison. The freque
of B is 256 Hz. When the prongs of A are scraped by
file and sounded with B, then 4 beats per second are
heard. What is the frequency of A after scrapping?​

Answer 1

Answer:

The answer is :

260Hz

Explanation:

The frequency of "A" increases when it is filed. Let "f" be its new frequency.

(frequency of A)- (frequency of B) = no. of beats

f - 256 = 4

f = 256 + 4 = 260Hz

Answer 2

Frequency of A after scrapping is 260 Hz .

Given :

Frequency of tuning forks B , $f_b=256\ Hz.$

We need to find frequency of tuning forks A , $f_a.$

Now,

When the prongs of A are scraped by file its frequency will increases.

Also , after filing it and sounded with B , it produces 4 beats .

Therefore , $f_a-f_b=4\\\\f_a=f_b+4=(256+4)\ Hz=260\ Hz.$

Frequency of A after scrapping is 260 Hz .

Hence, this is the required solution.

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