Question

Two strings 'X' and 'Y' of a guitar produces a beat frequency of 6 Hz. When the tension of the string 'Y' is increased, the beat frequency is found to be 4 Hz. If the frequency of string 'X' is 300 Hz, then the original frequency of string 'Y' is

• A. 296 Hz
• B. 294 Hz
• C. 310 Hz
• D. 304 Hz

Hint:

When dealing with beat frequencies, the absolute difference between the two frequencies gives the beat frequency. A small change in tension leads to a noticeable change in the frequency and hence the beat frequency.

Answer 1

The Correct Option is D

Step 1: Beat frequency formula.
The beat frequency \( f_{\text{beat}} \) between two sound waves with frequencies \( f_1 \) and \( f_2 \) is given by the formula:
\[ f_{\text{beat}} = |f_1 - f_2|. \]
In the given problem, the initial beat frequency is 6 Hz, and after increasing the tension in string 'Y', the beat frequency becomes 4 Hz.
Let the original frequency of string 'Y' be \( f_Y \). We are given that the frequency of string 'X' is \( f_X = 300 \, \text{Hz} \).

Step 2: Applying the initial beat frequency.
Initially, the beat frequency is 6 Hz, so:
\[ |300 - f_Y| = 6. \]
This gives two possible equations:
\[ 300 - f_Y = 6 \quad \text{or} \quad f_Y - 300 = 6. \]
Solving these equations:
- If \( 300 - f_Y = 6 \), we get \( f_Y = 294 \, \text{Hz} \).
- If \( f_Y - 300 = 6 \), we get \( f_Y = 306 \, \text{Hz} \).

Step 3: Applying the new beat frequency.
When the tension in string 'Y' is increased, the beat frequency becomes 4 Hz. So:
\[ |f_X - f_Y'| = 4, \]
where \( f_Y' \) is the new frequency of string 'Y'. Since the tension increases, the frequency of string 'Y' must be higher than \( 294 \, \text{Hz} \), so we use \( f_Y' = 304 \, \text{Hz} \). Therefore:
\[ |300 - 304| = 4 \quad \text{which is true}. \]

Step 4: Conclusion.
Thus, the original frequency of string 'Y' was \( 304 \, \text{Hz} \).
Final Answer:
The original frequency of string 'Y' is:
\[ \boxed{304 \, \text{Hz}}. \]