A particle is performing SHM about mean position O as shown in the figure given below. Its amplitude is 5 m and time period of oscillation is 4 s. The minimum time taken by the particle to move from point A to B is nearly

Options

A. 0.18 s
B. 0.20s
C. 0.15s
D. 0.24s

Answer

A

Explanation

The displacement equation of SHM is
`x=AsinomegatRightarrow sinomegat=(x)/(A)Rightarrowomegat=sin^(-1)((X)/(A))`
Substiuting the given values , we get
`omegat_(1)=sin^(-1)((3)/(5))=37^(@)(37xx(pi)/(180))rad`…….(i)
And `omegat_(2)=sin^(-1)((4)/(5))=53^(@)=(53xx(pi)/(180))rad`.....(ii)
Subracting eq.(i)from eq.(ii),we get
`omega(t_(2)-t_(1))=(53-37)(pi)/(180)`
` Rightarrow t_(2)-t_(1)=(1)/(omega)(53-37)(pi)/(180)=(T)/(2pi)(16)(pi)/(180)=(8T)/(180)=(8xx4)/(180)=(8)/(45)=0.18s`.