Verified answer

T = 2 pi * sqrt(L/g)

Assuming g = 9.8 m/s^2, (For a pendulum on the Earth,)

2 pi * sqrt(10/9.8) = 6.346975625940523 seconds. You can round this to 6.3 s.

The frequency is the reciprocal of the period, and so 1/6.346975625940523 = 0.157555355 Hz, or around 0.16 Hz.

Hope this helps!

The formula for determining the time period of a pendulum is... Tp = 2(pi)sqrt(L/g)

Here, Tp = time period, L = length of the pendulum, and g = acceleration due to gravity.

So, plug in the values that you know... Tp = 2(pi)sqrt(10m/(9.8m/s^2)) => Tp = 6.347 sec.

To find the frequency, just take the inverse of the time period (this gives you an answer in cycles per second or Hertz)

6.347^-1 = 0.158 Hz

I hope this helps!

Happy New Year!

For small plenty, the formula for the duration of a pendulum is: T = 2* pi * sqrt (L/g) Divide the size of the rope with the aid of the gravitational acceleration (9.8 m/sec^2), take the sq. root, and multiply with the aid of 2 pi. different than subsequently, you already comprehend the era and don't comprehend the size. Rearrange the equation to resolve for L T/(2 pi) = sqrt (L/g) T^2/(4 pi^2) = L/g gT^2/(4 pi^2) = L