NEET2010PhysicsWaves and SoundActual
A tuning fork of frequency 512 ~Hz makes 4 beats/seconds with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
Options
- A510 ~Hz
- B514 ~Hz
- C516 ~Hz
- D508 ~Hz
Correct answer
D. 508 ~Hz
Step-by-step solution
Suppose n_p= frequency of piano = ? (n_p T ) n _ f = frequency of tuning fork =512 ~Hz x = Beat frequency =4 beats / s , which is decreasing (4 2) after changing the tension of piano wire. Also, tension of piano wire is increasing so n_p Hence, n _ p - n _ f = x wrong n _ f - n _ p = x correjct n _ p = n _ f - x =512-4=508 ~Hz