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CGP EDU Academic Team
Published on: September 12, 2026
Two adjacent piano keys are struck simultaneously. The notes emitted by them have frequencies $\Pi_1$ and $\mathrm{II}_2$ . The number of beats heard per second is
$(a) \frac{1}{2} (n_1 - n_2) \quad (b) \frac{1}{2} (n_1 + n_2) \\ (c) n_1 \sim n_2 \quad (d) 2 (n_1 - n_2)$
Text Solution
Verified by ExpertsThe correct answer is:
A
To find the number of beats heard per second when two sound waves of different frequencies interfere, we use the formula:
Number of beats = |n1 - n2|, where n1 and n2 are the frequencies of the two notes.
Since the question involves calculating the beats heard per second, we need to account for the beat frequency which can be expressed as:
Beats per second = 1/2 |n1 - n2|.
Therefore, option (a) correctly represents this relation: 1/2(n1 - n2). Therefore, A.
Number of beats = |n1 - n2|, where n1 and n2 are the frequencies of the two notes.
Since the question involves calculating the beats heard per second, we need to account for the beat frequency which can be expressed as:
Beats per second = 1/2 |n1 - n2|.
Therefore, option (a) correctly represents this relation: 1/2(n1 - n2). Therefore, A.
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