The top of a skyscraper sways back and forth, completing 9 oscillation cycles in 1 minute. Find the period and frequency of its motion.

Short Answer

Expert verified
The period of the skyscraper's oscillation is approximately 6.7 seconds, and the frequency of its motion is approximately 0.15 Hz.

Step by step solution

01

Calculate the period

Initially, the period (T) of the oscillator is calculated. From the problem statement, it is known that the oscillator completes 9 cycles in 1 minute. The period is the time it takes for one complete cycle, so it's the total time divided by the number of cycles. This can be expressed with the following equation: \(T = \frac{\text{total time}}{\text{number of cycles}}\). The total time is 1 minute (which is 60 seconds), and the number of cycles is 9. Thus, the period is \(T = \frac{60 \text{ seconds}}{9} = 6.7 \text{ seconds}\).
02

Calculate the frequency

Next, calculate the frequency (f) of the oscillator. The frequency is the number of cycles per second. Frequency is the reciprocal of the period, and can be calculated with the following equation: \(f = \frac{1}{T}\). Using our previously calculated period value: \(f = \frac{1}{6.7 \text{ seconds}} = 0.15 \text{ Hz}\). The unit for frequency is Hz (Hertz), which represents cycles per second.

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