Question 283: Waves

Question

When two tuning forks are sounded together, 6 beats per second are heard. One of the fork is in unison with 0.70 m length of sonometer wire and another fork is in unison with 0.69 m length of the same sonometer wire. The frequencies of the two tuning forks are

Accepted Answer

Step 1: Frequency and Length Relation
The fundamental frequency of a stretched string (like a sonometer wire) is inversely proportional to its length if the tension and mass per unit length are constant.
So,
$ f \propto \frac{1}{L} $
Let the frequencies when the sonometer lengths are $L_1 = 0.70$ m and $L_2 = 0.69$ m be $f_1$ and $f_2$ respectively.
Step 2: Set Up the Ratio
$ \frac{f_1}{f_2} = \frac{L_2}{L_1} = \frac{0.69}{0.70} $
Let $f_1 < f_2$, as longer length corresponds to a lower frequency.
Step 3: Express Frequencies and Beat Frequency
Given, number of beats per second $= 6$
$ |f_2 - f_1| = 6 $
Let $f_1 = x$, so $f_2 = x + 6$.
Step 4: Substitute and Solve
Using the ratio:
$ \frac{x}{x+6} = \frac{0.69}{0.70} $
Cross-multiplied:
$ 0.70x = 0.69(x + 6) $
$ 0.70x = 0.69x + 4.14 $
$ 0.70x - 0.69x = 4.14 $
$ 0.01x = 4.14 $
$ x = \frac{4.14}{0.01} = 414 $
So,
$ f_1 = 414~	ext{Hz} $
$ f_2 = 414 + 6 = 420~	ext{Hz} $
Step 5: Match with Options
So, the correct answer is:
$ \boxed{414~	ext{Hz},~420~	ext{Hz}} $
Option B is correct.

Answer

$320 \mathrm{~Hz}, 326 \mathrm{~Hz}$

Answer

$414 \mathrm{~Hz}, 420 \mathrm{~Hz}$

Answer

$420 \mathrm{~Hz}, 426 \mathrm{~Hz}$

Answer

$480 \mathrm{~Hz}, 486 \mathrm{~Hz}$

Waves

When two tuning forks are sounded together, 6 beats per second are heard. One of the fork is in unison with 0.70 m length of sonometer wire and another fork is in unison with 0.69 m length of the same sonometer wire. The frequencies of the two tuning forks are

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