Question

A standard audio oscillator is used for a tuning fork. When the audio oscillator reads 514 Hz, two beats are produced in every second. When the audio oscillator reads 510 Hz then the beat frequency is 6 Hz. The required frequency of tuning fork is :
A) 506
B) 510
C) 516
D) 158

Answer 1

Hint: When two sounds waves of slightly different frequencies travelling along the same path in the same direction and same medium superpose upon each other, the intensity of the resultant sound at any point in the medium rises and falls (waxing and waning of sound) alternately with time. These periodic changes in the intensity of waves caused by $({v_o} - 514) = 2$ the superposition of two sound waves of different frequencies are called beats.

Complete step by step solution:
As we all already knows that:
The number of beats produced by the oscillator in every second is known as beat frequency.
Beat frequency = Difference in frequencies of the waves.
Mathematically, ${v_{beat}} = {v_1} - {v_2}$
where ${v_1}$ is frequency of first sound wave
${v_2}$ is frequency of second sound wave
So, difference in Beat frequency $v = ({v_o} - 514) = 2$
Also, $({v_o} - 510) = 6$
On solving both equations
We get beat frequency = $516 Hz$

Note: Necessary conditions for the production of beats:- For audible beats , the difference in frequencies of the two sound waves should not be more than 10, if the difference in frequencies is more than 10, we shall hear more than 10 beats per second. But due to persistence of hearing, our ear is not able to distinguish between the beats in less than (1/10) of a second. Hence beats to be heard will not be distinct if the number of beats produced per second is more than 10.