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Period of a pendulum (T) in Seconds is:
T = 2 * PI * (L/g)1/2
L = Length of Pendulum in Meters
g = Acceleration due to gravity = 9.81 m/s2
PI = 3.14
The period is independent of the mass or travel (angle) of the pendulum.
The frequency (f) of a pendulum in Hertz is the inverse of the Period.
f = 1/T
What else can I help you with?
The gravitational acceleration on the surface of Mars is 3.7m s 2 What period Would a 64 meters long pendulum have on Mars?
The period of a pendulum on Mars compared to Earth would be about 1.62 times longer.The period of a pendulum is (among other factors) inversely proportional to the square root of the acceleration due to gravity. The gravity of Mars is 0.38 that of Earth, so the square root of one over 0.38 is 1.62.T ~= 2 pi sqrt (L/g) where theta far less than 1.For larger theta, longer periods are incurred, with various correction factors, but the basic equation remains the same.
House current in the US oscillates how many times in one second?
The general oscilattion rating for a United States household electrical system is 60 times per second. This is know as Hrtz (prnounced Hets). Therefore, home electrical systems in the US run at 60 Hrtz.
In 6 seconds a light flashes once In one hour how many times it will flash?
600, 60(seconds)x60(minutes)=3600(seconds)/6=600
An electron in hydrogen atom stays in 2nd orbit for 10 -8 s how many revolutions will it make till it jumps to the ground state?
The time an electron spends in the second orbit of a hydrogen atom is (10^{-8}) seconds. The orbital period for an electron in the second orbit can be calculated using the formula for the orbital period in a hydrogen atom, which is approximately (T = 1.52 \times 10^{-16}) seconds. Therefore, the number of revolutions the electron makes before jumping to the ground state is approximately ( \frac{10^{-8}}{1.52 \times 10^{-16}} \approx 6.58 \times 10^7) revolutions.
Which is smallest period?
The smallest period in physics is often referred to as the Planck time, which is approximately (5.39 \times 10^{-44}) seconds. It represents the time it would take for light to travel one Planck length, the smallest measurable length, in a vacuum. At this scale, the effects of quantum gravity are believed to dominate, making it a fundamental unit in the study of the universe's earliest moments.
A pendulum oscillates 60 times in 6 seconds.Find its time period and frequency?
Frequency=60/6=10Hz Time Period=1/f=1/10
What is a period wave with the frequency of 2.6 Hz?
A period wave with a frequency of 2.6 Hz completes one full cycle (peak to peak or trough to trough) every 0.38 seconds. This means it oscillates 2.6 times in one second.
A pendulum that is 38 cm long will swing how many times in 15 seconds?
12.
How do you find determine the period of the pendulum A pupil obtains the following values for 20 oscillations of a pendulum 12.6s 12.7s 12.5s 12.6s 12.7s Determine the period of pendulum?
First take the average of your times:(12.6 + 12.7 + 12.5 + 12.6 + 12.7) / 5 = 12.62This is your average time for 20 oscillations. The period is the time for one oscillation, and therefore the period is 12.62/20 = 0.631 seconds.A complete oscillation is when the pendulum swings from the start position to the opposite position on the swing and back again. Assuming this is what you counted twenty of, then your pendulum is 10 cm long.If you counted 20 swings to each side, then you really only counted 10 oscillations. This means that your period would be 1.262, and would suggest that your pendulum is 40cm long.
Time period of a pendulum on moon is?
About 40.7% of that on Earth or about 2.46 times slower.
How does frequency affect the pendulum?
The frequency of a pendulum is related to its period, or the time it takes to complete one full swing. The frequency increases as the pendulum swings faster and the period decreases. In essence, an increase in frequency means the pendulum is swinging more times per unit of time.
What is the difference in period for a pendulum on earth and a pendulum on moon?
The period of a simple pendulum swinging at a small angle is approximately 2*pi*Sqrt(L/g), where L is the length of the pendulum, and g is acceleration due to gravity. Since gravity on the moon is approximately 1/6 of Earth's gravity, the period of a pendulum on the moon with the same length will be approximately 2.45 times of the same pendulum on the Earth (that's square root of 6).
How many times does a 32 inch pendulum swing in one minute?
Period ( left swing + right swing) of a simple pendulum = 2*pi * sqrt (L/g) in seconds. g = 32.2 feet per second2 L = 32 inches = 2.66667 feet Period = 2*pi * sqrt ( 2.66667ft/32.2) = 2*pi * 0.287777 = 1.808158 seconds for one period (two swings). Periods in one minute = 60 sec / 1.808158 sec = 33.183 periods in one minute. Times 2 = 66.366 swings in one minute.
What changes can you make to affect the pendulum so that it can swing more or less swings in 15 seconds?
To make the pendulum swing more times in 15 seconds, you can increase its length or increase the angle of release. To make it swing less in 15 seconds, you can decrease the length or reduce the angle of release. Additionally, reducing air resistance by swinging in a vacuum can also affect the number of swings in 15 seconds.
How does the length of a string affect the number Of times a pendulum will swing back and fourth in 10 seconds?
The length of a pendulum affects its period of oscillation, which is the time taken for one complete swing back and forth. A longer pendulum will have a longer period and therefore fewer swings in a given time period, like 10 seconds. Conversely, a shorter pendulum will have a shorter period and more swings in the same time frame.
If the length of a simple pendulum is doubled what will be the change in its time period?
ts period will become sqrt(2) times as long.
How many times does a 80 in pendulum swing in one minute?
The number of swings of a pendulum in one minute depends on its length and the acceleration due to gravity. For a simple pendulum, the period (time for one complete swing) can be calculated using the formula ( T = 2\pi \sqrt{\frac{L}{g}} ), where ( L ) is the length and ( g ) is the acceleration due to gravity (approximately 9.81 m/s²). An 80 cm (or 0.8 m) pendulum has a period of about 1.79 seconds, leading to roughly 33 swings in one minute.
