Question

A particle performs SHM between two points 0.2 m apart with a period 10 s. How far will the particle be in 1 s. after leaving the mid-position between two points ?

Ans. 58.8 mm)​

Answer 1

Answer:

Amplitude of particle executing SHM 0.2m. Time period T=24s

Hence 

x=0.2sinωt

ω is angular frequency equal to 242π

In question x=0.1m and we have to find time t

Putting the value of x

0.1=0.2sinωt

sinωt=21=sin6π

242πt=6π

⇒t=2s