A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is (a) 105 Hz (b) 1.05 Hz (c) 1050 Hz (d) 10.5 Hz

Answer: (a). Consecutive resonant frequencies differ by the fundamental. f₀ = 420 - 315 = 105 Hz ( 315 = 3f₀, 420 = 4f₀). So the answer is (a) 105 Hz.

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Stationary Waves & Vibrating Strings

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. Then, the lowest resonant frequency for this string is

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Answer: (a)

Consecutive resonant frequencies differ by the fundamental.
$f_0 = 420 - 315 = 105\text{ Hz}$ ( $315 = 3f_0$ , $420 = 4f_0$ ).
So the answer is (a) $105\text{ Hz}$ .

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