Wave motionMedium
Question
A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequncies of 420 Hz and 315 Hz. There are no other resonant frequenciens between these two. Then, the lowest resonant frequency for this string is
Options
A.105Hz
B.1.05Hz
C.1050 Hz
D.10.5 Hz
Solution
Given 
= 315 and (n + 1) 
= 420
⇒
⇒ n = 3
Hence 3 ×
= 315 ⇒ = 105 Hz
The lowest resonant freqency is when n = 1
therefore lowest resonant frequency = 105Hz.
= 315 and (n + 1) = 420 ⇒
⇒ n = 3Hence 3 ×
= 315 ⇒ = 105 Hz The lowest resonant freqency is when n = 1
therefore lowest resonant frequency = 105Hz.
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