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Q14.18: NCERT Physics - Part 2 WAVES 14.18 Solution: Two sitar strings A and B playing the note 'Ga' are slightly…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often fail to link the reduction in string tension to a decrease in frequency, leading them to incorrectly select the higher frequency option. Remember that $f \propto \sqrt{T}$, so reducing tension lowers $f_A$; if the beat frequency also decreases, $f_B$ must be lower than the original $f_A$ to maintain a smaller difference.

Two sitar strings A and B playing the note 'Ga' are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Step-by-Step Verified Solution

**Answer:** fB=318f_B = 318 Hz **Step-by-Step Explanation:** Beat frequency = ∣fA−fB∣=6|f_A - f_B| = 6 Hz. fA=324f_A = 324 Hz, so fB=318f_B = 318 Hz or 330330 Hz. When tension in A is reduced, its frequency decreases. If fB=330f_B = 330 Hz, reducing fAf_A would increase the beat frequency. But beat frequency reduces to 3 Hz, so fBf_B must be 318 Hz. Reducing fAf_A to 321 Hz gives beat frequency 321−318=3321 - 318 = 3 Hz [citation:13].

All Questions in this Chapter

Q14.1 Q14.2 Q14.3 Q14.4 Q14.5 Q14.6 Q14.7 Q14.8 Q14.9 Q14.10 Q14.11 Q14.12 Q14.13 Q14.14 Q14.15 Q14.16 Q14.17 Q14.18 Q14.19

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