Standing Wave in String and Organ Pipe: Practice Problem & Solution
A string is stretched between fixed points separated by $75.0 \text{ cm}$. It is observed to have resonant frequencies of $420 \text{ Hz}$ and $315 \text{ Hz}$. There are no other resonant frequencies between these two. The lowest resonant frequencies for this string is: (2015 Re)
Solution Explained:
To solve this problem, we apply the core principles of Standing Wave in String and Organ Pipe. Understanding the underlying formula is key to arriving at the correct answer below:
The lowest resonant (fundamental) frequency $f$ is the difference between two consecutive resonant frequencies. $f = f_{n+1} - f_n = 420 - 315 = 105 \text{ Hz}$.
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