Question:

If the resonance frequency of an acoustic system is 300Hz and the half-power frequencies are 150Hz and 450Hz, the quality factor is:

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By definition, half-power frequencies occur where the power drops to half its peak value, so f2 − f1 measures the bandwidth around f0. Consequently, Q = f0/bandwidth.
Updated On: Jan 6, 2025
  • 1.25
  • 1.5
  • 1
  • 1.75
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The Correct Option is C

Solution and Explanation

A common definition for the quality factor \( Q \) around a resonance \( f_0 \) is \[ Q = \frac{f_0}{f_2 - f_1}, \] where \( f_1 \) and \( f_2 \) are the half-power (or \(-3 \, \text{dB}\)) frequencies. Here: \[ f_0 = 300 \, \text{Hz}, \quad f_1 = 150 \, \text{Hz}, \quad f_2 = 450 \, \text{Hz}. \] Thus, \[ f_2 - f_1 = 450 \, \text{Hz} - 150 \, \text{Hz} = 300 \, \text{Hz}, \quad \Rightarrow \quad Q = \frac{f_0}{f_2 - f_1} = \frac{300}{300} = 1.0. \] Hence the quality factor is \fbox{1.0}.

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