Question

In a certain oscillatory system (particle is performing SHM), the amplitude of motion is 5 m and the time period is 4 s. the minimum time taken by the particle for passing betweens points, which are at distances of 4 m and 3 m from the centre and on the same side of it will approximately be
A. `(16)/(45)s`
B. `(7)/(45)s`
C. `(8)/(45)s`
D. `(13)/(45)s`

Answer 1

Correct Answer - C
`x=Asinomegat`, where `omega=(2pi)/(4)=(pi)/(2)(rad)/(s)`
`3=5sinomegat_1impliest_1=(1)/(omega)sin^-1((3)/(5))=(2xx37)/(180)`
`4=5sinomegat_2impliest_2=(1)/(omega)sin^-1((4)/(5))=(2xx53)/(180)`
Hence time interval `(t_2-t_1)=(1)/(90)(53-37)=(8)/(45)s`