Question

A particle executing SHM along a straight line has a velocity of `4ms^(-1)`, and at a distance of 3m from its mean position and `3ms^(-1)`, when at a distance of 4m from it. Find the time it take to travel 2.5m from the positive extremity of its oscillation.

Answer 1

Here, (case1), `V_(1)=4ms^(-1),y_(1)=3m,`
case(II), `V_(2)=3ms^(-1),y_(2)=4m.`
We know that, `V=omegasqrt(a^(2)-y^(2))`
Case (I) `4=omegasqrt(a^(2)-3^(2))`..(i)
Case(II) `3=omegasqrt(a^(2)-4^(2)` …(ii)
Dividing (i) by (ii) , we get
`(4)/(3)=(omega sqrt(a^(2)-9))/(omegasqrt(a^(2)-16)) or (16)/(9)=(a^(2)-9)/(a^(2)-16)`
or `16a^(2)-256=9a^(2)-81`
or `7a^(2)=256-81=175`
or `a^(2)=(175)/(7)=25 or a = sqrt(925)=5m`
Substituting it in (i), we get
`4=omegasqrt(5^(2)-3^(2))=omegasqrt(25-9)=omegaxx4 `
or `omega =4//4 = 1rads^(-1)`
When the particel is at a distance 2.5m from the extreme position, then its distance from the mean position, `x=5-2.5=2.5m. `
Since, the time is to be noted from the extremen position for SHM therefore, we shall use the relation
`x=acosomegat.`
or `2.5=5cos1xxt=5cost`
or `cost=(2.5)/(5)=(1)/(2)=cos((pi)/(3)) `
or ` t=(pi)/(3)=(22)/(7xx3)=1.048s`