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

1 Answer
May 25, 2017

#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#