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The weight (or mass) is irrelevant.

The period t = sqrt(2*pi*sqrt(l/g) where g is the acceleration due to gravity.

so t = 2*pi*sqrt(5/9.8) = 4.49 seconds.

Frequency = 1/period = 0.223 Hertz.

Q: 1 What is the period and frequency of a 10-pound weight oscillation a 5-meter-long rope?

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The period of a wave is defined as the time taken by a wave to complete one oscillation. While, the frequency of a wave is defined as the number of oscillations completed by a wave in one second.

Frequency = 1/period1/7.5 x 10-3 = 1331/3 Hz = 2/15 KHz

The relationship to be kept in mind is that the time period, which is the time take for one complete oscillation, is the reciprocal of frequency (the number of oscillations completed in one second). That is, T = 1/f where Period T, is measured in seconds Frequency f, is measured in Hertz. In your specific case, if f = 10 Hz, T = 0.1 s.

Period = 1 / frequency

Time period per oscillation=32/ 20=1.6 sec per oscillation.

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The inverse of frequency.

Time period and frequency are mutual reciprocals. T = 1/f F = 1/t

T=1/f .5=1/f f=2

The time taken by one complete oscillation to cross a fixed point is called the period of a wave.

The period of a wave is defined as the time taken by a wave to complete one oscillation. While, the frequency of a wave is defined as the number of oscillations completed by a wave in one second.

Assuming no friction, the period of oscillation would be the same if the length from the center of mass of the milk or bread to the pivot point is equal.

Frequency = 1/period1/7.5 x 10-3 = 1331/3 Hz = 2/15 KHz

Wouldn't that depend on what's doing the oscillating ? If it's a stone on the end of a rubber band, or a pendulum, or anything else like that where the period depends on the weight of something, then of course it's going to be different, because the weight of the weight is different. But if it's an oscillation of voltage and current in an electronic circuit, then weight/gravity has nothing to do with the frequency. The frequency of radio transmissions from astronauts in flight, as well as the frequency of their voices for that matter, is the same whether they're on Earth, on the moon, or weightless in orbit somewhere.

make the rod longer the rod will shorten the period. The mass of the bob does not affect the period. You could also increase the gravitational pull.

In order to find the frequency of an oscilloscope trace, you must first find the period, which is the time it takes for one oscillation, which can be found by measuring the amount of time from one peak our trough to the next. The frequency is the number of oscillations per second, and can be found by dividing 1 by the period in seconds.

This sounds like a homework question, so rather than doing your homework for you, I'll explain how to find the answer for yourself.The period and the frequency are very easy to find from the information provided. First, remember the definitions of period and frequency:Period is the amount of time per oscillation.Frequency is the number of oscillations per unit time.When you see the word "per" in these definitions, you should think division, because that is how you find the answer. When we say period is time per oscillation, this means time divided by oscillations. So your period is obtained by dividing the amount of time (in seconds) by the number of oscillations in that time. The units will make sense this way too: time divided by oscillations will give you a number that means seconds PER oscillation, which is what period is!Similarly, frequency is oscillations per time, so take the number of oscillations and divide it by the number of seconds it took. Your units will be oscillations per second, which makes sense for a measure of frequency.

Mass oscillation time period = 2 pi sq rt. (m/k) Pendulum oscillation time period = 2 pi sq rt. (l/g)