Question

A tuning fork of unknown frequency produces 4 beats per second when sounded with another tuning fork of frequency 254 hz

Answer 1

Answer:

The unknown frequency could be either 250Hz or 258Hz

Explanation:

Given : number of beats produced per second = 4

Frequency of the tuning fork = 254 Hz

We are needed to find the unknown frequency of the fork

 Let the unknown frequency be f.

Number of beats= change in frequency of the two forks

                  4=  254-f

                  f=250 Hz

this is one possibility

Other possibility

   4= f-254

  f=258 Hz

The unknown frequency could be either 250Hz or 258Hz

Answer 2

Answer:

258Hz

Explanation:

f1-f2 = 4

If f1= 254 Hz then f2= 250Hz

If f2=254 Hz then f1 = 258Hz

To solve this confusion,

The data given states if the unknown frequency is loaded with wax the beats remain constant.

Loading wax reduces frequency,

If it were 250Hz further decrease (say for example 249Hz)l,the beat frequency becomes 5 that is gets increased continuously.

If it were 258 Hz it can get reduced upto 254 where both are equal (beat frequency will decrease, check using values if any doubt)and consecutive decrease can cause 254 to be greater in magnitude and unknown to reach values less than 254.

Now the beat frequency increases continuously.

Therefore it can reach from 0 to4 .

Thus, 258 Hz is right.