1hey ashish if u have hc verma c page no.252 question no. 7
CONSIDER A PARTICLE MOVING IN SHM ACCORDING TO EQUATION x=2cos(50Ï€t+tan-10.75) x is in cdentimetre n t is in second motion starts at t=0 (A) WHEN DOES THE PARTICLE COMES AT REST FOR THE FIRST TIME? (B) WHEN DOES THE PARTICLE HAS MAXIMUM ACCELARATION FOR THE FIRST TIME? (C) WHEN DOES THE PARTICLE COMES AT REST FOR THE SECOND TIME?
-
UP 0 DOWN 0 0 7
7 Answers
106hey how can particle perform SHM by the equation u have given???? it cant..
62what is f it depends on that!
because there is no t...
so x is fixed..
so the body is at rest at all times!
106(I)
then particle will come to rest for first time when v = dx/dt = 0
i.e. -2sin(50Ï€t + than-10.75).50Ï€ = 0
==> sin(50Ï€t + than-10.75) = 0
==>50πt + than-10.75 = nπ
==> t=(nπ - tan-10.75)/50π
Calculate the min. valu of n for which t>0 i.e. n=1
So, t=(Ï€-tan-10.75)/50Ï€
106(III) when it comes to rest for 2nd time n=2 from above equation..
So, t=(2Ï€-tan-10.75)/50Ï€
106For (I) and (III) put π=180° and solve.... ans will come and tan-10.75 as 37°
(II)
a = dv/dt = d(-100Ï€sin(50Ï€t+tan-10.75)/dt
= -500Ï€2cos(50Ï€t+tan-10.75)
using concept of maxima and minima,
amax is when da/dt =0
==> 25000Ï€3sin((50Ï€t+tan-10.75) = 0
==> sin((50Ï€t+tan-10.75)=0
this is same as (I) equation
hence t= (Ï€-tan-10.75)/50Ï€
or this can be solved by.... a particle in SHM has maxm acceleration when its speed is minm (here 0) so, it is answer of (I) only....