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NEET PHYSICSMedium

A string is stretched between fixed points separated by 75.0 cm75.0 \text{ cm}. It is observed to have resonant frequencies of 420 Hz420 \text{ Hz} and 315 Hz315 \text{ Hz}. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is:

A

155 Hz155 \text{ Hz}

B

205 Hz205 \text{ Hz}

C

10.5 Hz10.5 \text{ Hz}

D

105 Hz105 \text{ Hz}

Step-by-Step Solution

  1. Identify Resonant Frequencies for a Fixed String: For a string fixed at both ends, the resonant frequencies are integer multiples (harmonics) of the fundamental frequency, given by fn=nf1f_n = n f_1, where n=1,2,3,…n = 1, 2, 3, \dots and f1f_1 is the fundamental (lowest) resonant frequency .
  2. Set up the Equations: The problem states there are no other resonant frequencies between 315 Hz315 \text{ Hz} and 420 Hz420 \text{ Hz}, which means they are consecutive harmonics. Let them be the nn-th and (n+1)(n+1)-th harmonics: fn=nf1=315 Hzf_n = n f_1 = 315 \text{ Hz} fn+1=(n+1)f1=420 Hzf_{n+1} = (n+1) f_1 = 420 \text{ Hz}
  3. Calculate the Fundamental Frequency: The difference between any two consecutive harmonic frequencies of a string fixed at both ends is equal to the fundamental frequency: fn+1−fn=(n+1)f1−nf1=f1f_{n+1} - f_n = (n+1)f_1 - n f_1 = f_1 f1=420 Hz−315 Hz=105 Hzf_1 = 420 \text{ Hz} - 315 \text{ Hz} = 105 \text{ Hz} Therefore, the lowest resonant frequency (fundamental frequency) is 105 Hz105 \text{ Hz}.
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