Question

A particle is moving with SHM along a straight line when the distance of the particle from the mean position is 3m and 4m the corresponding values of velocities are 4m/s and 3m/s. what time it will take to travel 2m from the positive extremity of the oscillation​

Answer 1

Answer:

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

I don't know this answer is right or not if it is wrong so sorry