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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