Question

A child swings with a small amplitude on a playground with a 2.5m long chain.
a) what is the period of the child’s motion
b) what is the frequency of vibration

Answer 1

a) T=2*pi(L/g)^1/2
T=2*3.14(2.5/9.81)^1/2
T=3.171s
b) T=1/f
f=1/3.171
f=0.315Hz

Answer 2

a. The period of the child's motion is 3.171 seconds.

b. The frequency of the vibration is 0.315 Hz.

Calculation of the period & frequency:

Since A child swings with a small amplitude on a playground with a 2.5m long chain.

a. So here the time period should be

$2\times \ pi(L/g)^{1/2}\\\\2\times 3.14(2.5/9.81)^{1/2}$

T=3.171s

b) Now the frequency is

$T=1\div f\\\\f=1\div 3.171$

f=0.315Hz

learn more about frequency here: https://brainly.com/question/22971850