Question 265: Waves

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

Accepted Answer

Given:
String $X$ has frequency $f_X = 300 \, 	ext{Hz}$.
String $Y$Ã¢ÂÂs frequency is initially $f_Y$ (unknown).
Beat frequency initially = $6 \, 	ext{Hz}$.
That means $|f_Y - f_X| = 6$.
So, $\; f_Y = 300 \pm 6 = 306\ 	ext{Hz or } 294\ 	ext{Hz}$.
Then the tension of string $Y$ is increased, so its frequency
increases
.
New beat frequency = $4 \, 	ext{Hz}$.
Case 1: $f_Y = 306 \, 	ext{Hz}$.
If we increase tension, the frequency rises above 306.
The difference with 300 will be larger than 6, so the beat frequency would
increase
, not decrease.
This does not match.
Case 2: $f_Y = 294 \, 	ext{Hz}$.
Beat frequency initially = $|300 - 294| = 6$. Ã¢ÂÂ Correct.
Now as tension increases, YÃ¢ÂÂs frequency increases above 294, moving towards 300.
So the difference reduces: $|f_Y - 300| = 4$.
That matches the "4 Hz" condition. Ã¢ÂÂ
Thus, the original $f_Y = 294 \, 	ext{Hz}$.
Ã¢ÂÂ
Correct Answer: Option B (294 Hz).

Answer

296 Hz

Answer

294 Hz

Answer

306 Hz

Answer

304 Hz

Waves

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

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