Practice question
Question
Two strings produce beats of 6 Hz. One has a frequency of 300 Hz. When the tension in the second string
is decreased, the beat frequency becomes 4 Hz. What was the original frequency of the second string?
Explanation
**Beat formation** is interference in time with time-varying amplitude. Frequencies close together generate slow modulation, count in given duration obtained by multiplying beat frequency by duration, e.g., 6 Hz × 5 s = 30 beats. Let v₂ be the original frequency. |300 - v₂| = 6 ⇒ v₂ = 294 Hz or 306 Hz . Decreasing tension decreases frequency. If v₂ = 306 , new v₂’ < 306 , beat = 306 - 300 = 6 or decreases, so v₂’ = 296 , beat = 300 - 296 = 4 , consistent. Thus, v₂ = 306 Hz . Using v = fλ and standing-wave condition
Discussion
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