Question:medium

Two identical piano wires, when tuned to a fundamental frequency of 400 Hz, produce no beats. One wire is then slightly tightened, and the beat frequency heard is 2 Hz. What is the new fundamental frequency of the tightened wire?

Show Hint

Tighten = Frequency Up. Loosen = Frequency Down.
  • 398 Hz
  • 402 Hz
  • 404 Hz
  • 396 Hz
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Beat frequency is the absolute difference between two frequencies (\( f_b = |f_1 - f_2| \)). The fundamental frequency of a stretched string is directly proportional to the square root of its tension (\( f \propto \sqrt{T} \)).
Step 2: Key Formula or Approach:
1. \( f_{new} = f_{old} \pm f_{beat} \).
2. If tension increases, frequency increases.
Step 3: Detailed Explanation:
Initial frequency \( f_1 = 400 \text{ Hz} \). The wire is tightened, so the new frequency \( f_2 \) must be greater than \( f_1 \). Given beat frequency \( f_b = 2 \text{ Hz} \): \[ f_2 - f_1 = 2 \] \[ f_2 - 400 = 2 \implies f_2 = 402 \text{ Hz} \]
Step 4: Final Answer:
The new fundamental frequency of the tightened wire is 402 Hz.
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