Question

What is the period of a pendulum near Earth's surface that is 130 cm long?

Answer 1

`T~~2.2876585681s`

Explanation:

Since you were not given the degrees of amplitude or radians of amplitude, we assume that the degrees of amplitude are less than `20^@` or `20^@` and we also assume that this is a simple pendulum

Apply the equation

`T ~~ 2pisqrt(L/g)" "(1)`

And for accurate

`T = (2pi)/("amg"(1,cos((pi * "degrees")/360^@)))xxsqrt(L/g)" "(2)`

Where `g` is acceleration due to gravity that is 9.80665m/s

`"amg(arithmetic geometric mean)"` is an operation

T is the time period

Time period is the time required for one oscillation

Where L is length

But we would use equation one

`T ~~ 2pisqrt("1.3m"/"9.80665m/s")`

The unit for length here is m

`T~~2.2876585681s`