# Answer to Question #42023 in Mechanics | Relativity for qwerty

Question #42023

A NUMBER OF TUNING FORKS ARE ARRANGED IN ORDER OF INCREASING FREQUENCY AND ANY 2 SUCCESSIVE TUNING FORKS PRODUCE 4BEATS/S WHEN SOUNDED TOGETHER. IF THE LAST TUNING FORK HAS A FREQUENCY OCTAVE HIGHER THAN THE FIRST AND FREQUENCY OF THE FIRST IS 256 Hz. THEN THE NUMBER OF TUNING FORKS IS :

1. 63

2. 64

3. 65

4. 66

1. 63

2. 64

3. 65

4. 66

Expert's answer

The frequency of the last tuningfork octave higher than the first one of a frequency 226 Hz. So, it equals 512

Hz. Thus, the number of tuning forces is

(512-256)/4 = 64.

Answer is 2. 64 tuning forces.

Hz. Thus, the number of tuning forces is

(512-256)/4 = 64.

Answer is 2. 64 tuning forces.

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

Assignment Expert14.10.18, 20:43Dear visitor,

please use panel for submitting new questions

Prashant12.10.18, 10:08Sir send the full solution.... that how we get 226 hz.. in your ans.

## Leave a comment