Question

The fequency of tuning fork A and B are 2% more and 2% less than the frequency of tuning forck C respectively . when A and B are simulataneously excited , 8 bites per second are production then the frequency of B will be

Answer 1

Answer:

196 hertz

Explanation:

Given The frequency of tuning fork A and B are 2% more and 2% less than the frequency of tuning fork C respectively . when A and B are simultaneously excited , 8 beats per second are produced then the frequency of B will be

Let the frequency of the tuning fork be t. The beats between A and B = f A and f B is 8 beats per second. Given A is 2% more and B is 2% less than C.  

So A will be 100 + 2 = 102 hertz and B will be 100 – 2 = 98 hertz.

Now f A = 102 / 100 x t and f B = 98 / 100 x t

So f A – f B = 102 / 100 t – 98 / 100 t

                     = 102 t – 98 t / 100

                    = 4 t / 100

      Now f A – f B  = t / 25

     Given f A – f B = 8

                t / 25 = 8

              t = 200 hertz

             Now to find frequency of B, so  

       f B = 98 / 100 x 200 = 196 hertz

   So frequency of B = 196 hertz

Answer 2

Answer:

The frequency of B is 156.8 Hz.

Explanation:

Given that,

Let n be the frequency of fork C

The frequency of A is

$n_{A}=n+\dfrac{3n}{100}$

$n_{A}=\dfrac{103n}{100}$....(I)

The frequency of B is

$n_{B}=n-\dfrac{2n}{100}$

$n_{B}=\dfrac{98n}{100}$.....(II)

We need to calculate the value of n

Using formula of frequency

$n_{A}-n_{B}=8$

Put the value into the formula

$\dfrac{103n}{100}-\dfrac{98n}{100}=8$

$\dfrac{5n}{100}=8$

$n=\dfrac{800}{5}$

$n=160\ Hz$

We need to calculate the frequency of B

From equation (II)

$n_{B}=\dfrac{98n}{100}$

$n_{B}=\dfrac{98\times160}{100}$

$n_{B}=156.8\ Hz$

Hence, The frequency of B is 156.8 Hz.