Question

A source of frequency n gives 5 beats per seconds when sounded with a source of frequency 200 Hz. The second harmonic frequency 2n of source gives 10 beats/s. when sounded with a source of frequency 420 Hz, the value of n is?

Answer 1

Hint:Here the formula of beat can be used to get the correct answer for this problem. The concept of beat can be used to calculate the final answer for this problem. The beat concept is used when there are two sounds of different frequency approaches the ears of any observer then there is formation of constructive and destructive interference takes place.

Formula used: The formation of the beat can be given as $f' = \left| {{f_1} - {f_2}} \right|$ where $f'$ is the resulting beat also ${f_1}$ and ${f_1}$ are the frequencies of the two sounds.

Complete step by step answer:
It is given that the source for frequency n gives a beat of 5 beats per second with the sound of frequency 200 Hz.
Therefore according to the concept of beat,
$\left| {n - 200} \right| = 5$………eq. (1)
Also the source of frequency 2n gives beat of 10 beats per second with a sound of frequency of 420 Hz so again applying the concept of beat we get,
$\left| {2n - 420} \right| = 10$………eq. (2)
Now if n is greater than 200 then,
$\
  n - 200 = 5 \\
  n = 205Hz \\
\ $
And if we check equation (2) for n=205 Hz. So we observe that the n=205 satisfies equation (2).
Now, if n is less than 200 Hz then,
$\
   - \left( {n - 200} \right) = 5 \\
  n = 195Hz \\
\ $
And if we observe the value of n=195 Hz then we see that equation is not being satisfied. Therefore the correct answer for this problem is $n = 205Hz$.

Note: The students should always remember the formula and concept of beat because for solving these types of problems the formula and concept of beat is must. The beat concept says that due to two or more sound the frequency of sound which is heard is affected.