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

Step 1: Understanding the Question:
Beats occur when two sound waves of nearly equal frequencies overlap. The beat frequency is the absolute difference between the two source frequencies.
Tightening a wire increases the tension. Frequency of a vibrating string is directly proportional to the square root of tension (\(f \propto \sqrt{T}\)).
Therefore, tightening the wire must increase its frequency.
Step 2: Key Formula or Approach:
Initial frequency \(f_1 = f_2 = 400 \text{ Hz}\).
Beat frequency \(f_b = |f_2' - f_1|\).
Given \(f_b = 2 \text{ Hz}\).
Decide whether \(f_2'\) is \(f_1 + 2\) or \(f_1 - 2\) based on the effect of tightening.
Step 3: Detailed Explanation:

We are given \(f_1 = 400 \text{ Hz}\).

Initially, no beats are heard, so the second wire also has \(f_2 = 400 \text{ Hz}\).

The second wire is tightened. According to the formula for frequency of a stretched string:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

Here, \(T\) is tension. Increasing \(T\) leads to an increase in frequency \(f\).

So the new frequency \(f_2'\) must be greater than 400 Hz (\(f_2'>400\)).

The beat frequency is given as 2 Hz:
\[ |f_2' - 400| = 2 \]

This implies \(f_2' = 400 + 2 = 402 \text{ Hz}\) OR \(f_2' = 400 - 2 = 398 \text{ Hz}\).

Since tightening increases frequency, the only valid answer is 402 Hz.

Step 4: Final Answer:
Tightening increases tension and thus the frequency. With a beat frequency of 2 Hz relative to 400 Hz, the new frequency is 402 Hz.