Question:
Three tuning forks A, B and C have respective frequencies \( n_1, n_2 \) and \( n_3 \) related as \( n_1 = 1.03n_2 \) and \( n_3 = 0.99n_2 \). When A and C are sounded together 4 beats are heard per second. The frequencies of fork B and C are respectively
Three tuning forks A, B and C have respective frequencies \( n_1, n_2 \) and \( n_3 \) related as \( n_1 = 1.03n_2 \) and \( n_3 = 0.99n_2 \). When A and C are sounded together 4 beats are heard per second. The frequencies of fork B and C are respectively
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- Beats = frequency difference
- Convert % relation into algebra before solving
Updated On: May 4, 2026
- 103 Hz and 100 Hz
- 103 Hz and 99 Hz
- 99 Hz and 100 Hz
- 100 Hz and 99 Hz
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The Correct Option is D
Solution and Explanation
Concept:
Beat frequency: \[ \text{beats} = |n_1 - n_3| \]
Step 1: Use given relations.
\[ n_1 = 1.03n_2,\quad n_3 = 0.99n_2 \]
Step 2: Apply beat condition.
\[ |n_1 - n_3| = 4 \] \[ |1.03n_2 - 0.99n_2| = 4 \]
Step 3: Simplify.
\[ 0.04n_2 = 4 \Rightarrow n_2 = 100\ \text{Hz} \]
Step 4: Find $n_3$.
\[ n_3 = 0.99 \times 100 = 99\ \text{Hz} \]
Step 5: Final answer.
\[ (n_2, n_3) = (100,\ 99) \]
Beat frequency: \[ \text{beats} = |n_1 - n_3| \]
Step 1: Use given relations.
\[ n_1 = 1.03n_2,\quad n_3 = 0.99n_2 \]
Step 2: Apply beat condition.
\[ |n_1 - n_3| = 4 \] \[ |1.03n_2 - 0.99n_2| = 4 \]
Step 3: Simplify.
\[ 0.04n_2 = 4 \Rightarrow n_2 = 100\ \text{Hz} \]
Step 4: Find $n_3$.
\[ n_3 = 0.99 \times 100 = 99\ \text{Hz} \]
Step 5: Final answer.
\[ (n_2, n_3) = (100,\ 99) \]
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