Two identical tuning forks vibrate at 587 Hz.After a small piece of clay is placed on one of them,eight beats per second are heard.What is the period of the tuning fork that holds the clay? A)1.68 × 10^-³ s B)1.70 × 10^-³ s C)1.73 × 10^-³ s D)1.76 × 10^-³ s E)1.80 × 10^-³ s

Question

Quizplus · Accepted Answer

The frequency of the tuning fork with the clay is either 587 Hz + 8 Hz = 595 Hz or 587 Hz - 8 Hz = 579 Hz. The period $T$ is given by $T = \frac{1}{f}$. For 595 Hz, $T = \frac{1}{595} \approx 1.68 \times 10^{-3} s$, and for 579 Hz, $T = \frac{1}{579} \approx 1.73 \times 10^{-3} s$. Since 1.68 × 10^-3 s corresponds to the unaltered tuning fork and 1.73 × 10^-3 s to the one with clay (because adding clay lowers the frequency), the correct answer is 1.73 × 10^-3 s.

Two Identical Tuning Forks Vibrate at 587 Hz