When a tuning fork of frequency 256 Hz is sounded together with unknown tuning fork, 4 beats are heard in one second. The frequency of the unknown tuning fork can be
Show Hint
Always use the \(\pm\) sign for unknown frequency problems unless there is extra information (like "frequency increases on waxing" or "decreases on filing").
Step 1: Understanding the Concept:
Beats are produced when two sound waves of slightly different frequencies interfere. The beat frequency (\(n\)) is equal to the absolute difference between the two frequencies. Step 2: Key Formula or Approach:
\[ n = |f_1 - f_2| \]
or \(f_2 = f_1 \pm n\). Step 3: Detailed Explanation:
Given:
Known frequency \(f_1 = 256 \text{ Hz}\).
Beat frequency \(n = 4 \text{ beats per second}\).
The frequency of the unknown fork (\(f_2\)) can be either higher or lower than the known frequency by the value of the beat frequency.
Case 1 (Higher): \(f_2 = 256 + 4 = 260 \text{ Hz}\).
Case 2 (Lower): \(f_2 = 256 - 4 = 252 \text{ Hz}\). Step 4: Final Answer:
The frequency can be 260 Hz or 252 Hz.