Question #72b8f

1 Answer
P dilip_k
Dec 8, 2017

Given that the amplitude of SHM is #a=10cm# and its time period is #T=4s#. So the particle executing SHM completes one oscillation in #4s# and 2 oscillations in #8s#
In one complete oscillation it covers 4 times the the length of its amplitude i.e. #4xx10cm=40cm#

So in #8s# the particle will cover #2xx40cm=80cm# distance

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