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Question:

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. The lowest resonant frequency for this string is:

155 Hz

10.5 Hz

205 Hz

105 Hz

Solution:

Here,
315 = n(ν/2L)
and 420 = (n+1)(ν/2L)
So, 420/315 = (n+1)/n ⇒ n = 3
Lowest resonant frequency will be (1/3) of 315 = 315/3 = 105 Hz