Question

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?

Answer 1

I would assume that if you place a piece of clay on one of the tuning forks, it would dampen the rate of vibration and hence have a lower vibrational frequency. With this in mind, a beat frequency is defined as the dissonant tone heard between two frequencies very close in tone. A beat frequency of 8 Hz then implies that the clay would be vibrating with a frequency of 579 Hz.

Having established the frequency of vibration, we also need to know the relationship between period and frequency which is expressed as \[T = \frac{ 1 }{ f }\]
Where T is the period of oscillation. By using this relationship we find that\[T = \frac{ 1 }{ 579 } = 0.00173Hz ^{-1} = 0.00173 s\]
A Hertz (Hz) is the same as 1/s so 1/Hz should give us the units to be "s" (Which it does). Hope this helps.