Question

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

• A. \(398\ \text{Hz}\)
• B. \(402\ \text{Hz}\)
• C. \(404\ \text{Hz}\)
• D. \(396\ \text{Hz}\)

Hint:

Tightening a string increases its frequency. So choose the frequency greater than the original frequency.

Answer 1

The Correct Option is B

Concept: Beat frequency is the difference between two frequencies: \[ f_b=|f_1-f_2| \]

Step 1: Initially, both wires have the same frequency: \[ f_1=400\ \text{Hz} \] So no beats are heard.

Step 2: One wire is slightly tightened.
When a wire is tightened, its tension increases. For a stretched string: \[ f\propto \sqrt{T} \] So if tension increases, frequency also increases.

Step 3: Beat frequency is given as: \[ f_b=2\ \text{Hz} \] Let the new frequency of tightened wire be \(f_2\). \[ |f_2-400|=2 \]

Step 4: Since the wire is tightened, frequency increases. \[ f_2=400+2 \] \[ f_2=402\ \text{Hz} \] Therefore, \[ \boxed{402\ \text{Hz}} \]