MCAT Physics Question 91: Answer and Explanation

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Question: 91

1. An opera singer has two precisely identical glasses. The singer produces as pure a tone as possible and shatters the first glass at a frequency of 808 Hz. She then sings a frequency of 838 Hz in the presence of the second glass. The second glass will likely:

  • A. shatter after a longer amount of time because the applied frequency is higher.
  • B. shatter after a shorter amount of time because the applied frequency is higher.
  • C. not shatter because the applied frequency is not equal to the natural frequency of the glass.
  • D. not shatter because higher-frequency sounds are associated with more attenuation.

Correct Answer: C

Explanation:

If these two glasses are perfectly identical, then the fact that the first glass shattered at 808 Hz tells us that this is very close (if not identical) to the natural (resonant) frequency of the glass. If she produces a frequency that is not equal (or very close) to the natural frequency, then the applied frequency will not cause the glass to resonate, and there will not be the increase in wave amplitude associated with resonating objects. Attenuation will increase with increased frequency because there is more motion over which nonconservative forces can damp the sound wave; however, even if sound level was matched to that which shattered the first glass when accounting for attenuation, the glass would still not shatter for the reasons described above, eliminating choice (D).

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