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A string is stretched between two 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 :
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- 205 Hz
- 10.5 Hz
- 105 Hz
- 155 Hz
Correct Option: C
In a stretched string all multiples of frequencies can be obtained i.e., if fundamental frequency is n then higher frequencies will be 2n, 3n, 4n ...
So, the difference between any two successive frequencies will be 'n'
According to question, n = 420 – 315 = 105 Hz
So the lowest frequency of the string is105 Hz.
