Question

Two tuning forks A and B produce 6 beats per second. Frequency of A is 300 H_(Z) . When B is slightly loaded with wax, beat frequency decreases . Find original frequency of B .

Answer 1

The frequency of B tuning fork is 294 Hz .

Explanation:

Given :

No. of beats $= 6$

Frequency of A tuning fork $f_{a} = 300$ Hz

Beats is nothing but the difference in frequency.

When wax is loaded the actual frequency of fork is reduced.

   $f_{a} - f_{b} =$ No. of beats

   $f_{b} = 300 -6$

   $f_{b} = 294$ Hz

Thus, the frequency of B tuning fork is 294 Hz

Answer 2

The original frequency of B is 306 Hz.

Explanation:

Beat frequency is given as,

$f_{beat} =f_{B} -f_{A}$

Here, $f_A$ is the frequency of tuning fork A and $f_{B}$ is the frequency of tuning fork B.

Given, $f_{beat} =6 beats/ s$ and $f_A=300 Hz$.

When B is loaded with wax, beat frequency will decrease,

so frequency of B is higher than that of A.

substitute the values, we get

$f_{B}-300 Hz=6$

$f_{B}=300 +6 = 306 Hz$

     = 306 Hz.

Thus, the original frequency of B is 306 Hz.

#Learn More: Beat frequency.

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