The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm s^-1 and 50 cms^-2. Find the position of the particle when the speed is 8cms^-1.

Answer

A, B, C

Explanation

`V_(maxO)=10cm/sec`
`rarr romega=10`……….1
`rarr omega^2=100/r^2`
`a_(max)=omega^2r50cmsec^-1`
`rarr omega^2=50/r=50/r`……..2
`:. 100/r^2=50/r`
` rarr r=2cm`
`:. omega=sqrt(100/r^2)=5sec^-1`
Again to find out the positions where the speed is 8 m/sec.
`v^2=w^2(r^2-y^2)`
`rarr 64=25(4-y^2)`
`rarr 64/25/4-y^2`
`4-y^2=2.56`
`rarr y^2=1.44`
`=y=sqrt(1.44)`
`rarr y=+-1.2xcm`
from mean position