Question

The displacement of a particle executing S.H.M is given by x=0.01sin100π(t+005).The time period is_______.​

Answer 1

Given:x=0.01\sin100\pi\left(t+0.05\right)=0.01\sin{\left(100\pi t+5\right)}x=0.01sin100π(t+0.05)=0.01sin(100πt+5)

S.H.M equation is x=A\sin{\left(\omega t+\phi\right)}x=Asin(ωt+ϕ)

Comparing the above equations, we get

A=0.01A=0.01, \omega t=100\pi tωt=100πt

\Rightarrow \omega=100\pi⇒ω=100π

\Rightarrow \dfrac{2\pi}{T}=100\pi⇒T2π=100π

\Rightarrow \dfrac{1}{T}=50⇒T1=50

\Rightarrow T=\dfrac{1}{50}\times \dfrac{2}{2}=\dfrac{2}{100}=0.02⇒T=501×22=1002=0.02 secs