Question

A source of frequency \(f\) gives 5 beats/sec when sounded with a source of frequency 200 Hz. The second harmonic of \(f\) gives 10 beats/sec when sounded with a source of frequency 420 Hz. The value of \(f\) is:

• A. 195 Hz
• B. 205 Hz
• C. 190 Hz
• D. 210 Hz
• E.

Hint:

When you have multiple choices for the frequency from the first set of beats, always calculate the harmonic for each and see which one produces the required beat frequency with the second source. It acts as an automatic verification.

Answer 1

The Correct Option is B

To determine the value of frequency \( f \), let's analyze the information and apply the principles of beats in sound waves:

1. Understanding Beats: 
   Beats occur when two different frequencies sound together. The beat frequency is the absolute difference between these two frequencies. 
   The formula for the beat frequency is given by: 
   \(|f_1 - f_2| = \text{Beat frequency}\)
2. Given Information:
   • The frequency \( f \) gives 5 beats/sec when paired with a 200 Hz source: 
     \(|f - 200| = 5\)
   • The second harmonic of \( f \) gives 10 beats/sec when paired with a 420 Hz source: 
     The second harmonic of \( f \) is \( 2f \). 
     \(|2f - 420| = 10\)
3. Solving the Equations:
   • Equation 1: \(|f - 200| = 5\)
     • Case 1: \( f - 200 = 5 \rightarrow f = 205 \)
     • Case 2: \( 200 - f = 5 \rightarrow f = 195 \)
   • Equation 2: \(|2f - 420| = 10\)
     • Case 1: \( 2f - 420 = 10 \rightarrow 2f = 430 \rightarrow f = 215 \)
     • Case 2: \( 420 - 2f = 10 \rightarrow 2f = 410 \rightarrow f = 205 \)
4. Conclusion: 
   To satisfy both beat conditions simultaneously, \( f = 205 \) is the valid solution that works for both equations without contradiction.

Therefore, the value of \( f \) is 205 Hz.