Question

derive an expression for period of a simple pendulum

Answer 1

You need to remember the formula of constant of simple pendulum such that:

k = L/T^2

You need to consider the value of constant equivalent to g/(4pi^2) (g expresses the gravity acceleration)

You need to set the equations g/(4pi^2) and L/T^2 equal such that:

L/T^2 = g/(4pi^2)

You need to find time period such that:

g*T^2 = 4pi^2*L

T^2 = (4pi^2*L)/g => T = 2pi*sqrt(L/g)

Hence, evaluating the time period of simple pendulum under given conditions yields T = 2pi*sqrt(L/g).
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Answer 2

To calculate the period of the pendulum, we can just plug in the given length into the equation above.

TTT=2πlg−−√=2π1m9.8m/s2−−−−−−−−√=2s

To find the amplitude, we'll use the equation given in the Period and Frequency lesson that gives us the velocity as a function of time. Since the problem says that the given velocity is the maximum velocity, we know that the pendulum is at the bottom of it's arc and 1/4th (or 3/4th's) of it's way through one period. Based on this knowledge, we can plug in 1/4 of the period for the change in time. We also know the frequency because we just found the period, so all we have to do is solve for the amplitude.

v(t)vmaxAAA=−2πfAcos(2πfΔt)=−2πfAcos(2πf14T)=−vmax2πfcos(2πf14T)=−2m/s2π12Hz∗cos(2π∗12Hz∗14∗2s)=.63m

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