A source of frequency \( \nu \) gives 6 beats/second when sounded with a source of frequency 200 Hz. The second Harmonic of frequency \( 2\nu \) of the source gives 8 beats/second when sounded with a source of frequency 420 Hz. The value of \( \nu \) is
Show Hint
Always solve beat frequency equations carefully and check for common values when multiple conditions are given.
Step 1: Understanding the Question:
A source of frequency \( \nu \) produces 6 beats per second with a 200 Hz source.
Its second harmonic \( 2\nu \) produces 8 beats per second with a 420 Hz source. Step 2: Key Formula or Approach:
Number of beats per second:
\[
\text{Beats} = |f_1 - f_2|
\]
Step 3: Detailed Explanation:
From the first condition:
\[
|\nu - 200| = 6
\]
\[
\nu = 200 \pm 6 = 206 \text{ Hz or } 194 \text{ Hz}
\]
From the second condition:
\[
|2\nu - 420| = 8
\]
\[
2\nu = 420 \pm 8 = 428 \text{ or } 412
\]
\[
\nu = 214 \text{ Hz or } 206 \text{ Hz}
\]
Now, the common value from both conditions is:
\[
\nu = 206 \text{ Hz}
\]
Step 4: Final Answer:
\[
\boxed{206 \text{ Hz}}
\]