Home
Class 11
PHYSICS
A particle performs SHM about origin alo...

A particle performs SHM about origin along X -axis with amplitude A. At t=0 it was located at `x=A/2` and going towards mean position with speed v .Find equation of SHM
(A) `y=A cos((2vt)/(sqrt(3))+(pi)/(3))`
(B)`y=A sin((2vt)/(sqrt(3)A)+(5 pi)/(6))`
(C) both A&B
(D) `y=A sin((2vt)/(A)+(5 pi)/(6))`

Promotional Banner

Similar Questions

Explore conceptually related problems

A particle executing SHM along y-axis, which is described by y = 10 "sin"(pi t)/(4) , phase of particle at t =2s is

sin^-1 (cos(sin^-1(sqrt(3)/2))= (A) pi/3 (B) pi/6 (C) - pi/6 (D) none of these

The value of cos^(-1)(cos((5pi)/3))+sin^(-1)(sin((5pi)/3)) is (a) pi/2 (b) (5pi)/3 (c) (10pi)/3 (d) 0

Two SHM are represcnted by equations y_(1)=6cos(6pit+(pi)/(6)),y_(2)=3(sqrt(3)sin3pit+cos3pit)

cos^-1 {-sin((5pi)/6)}= (A) - (5pi)/6 (B) (5pi)/6 (C) (2pi)/3 (D) -(2pi)/3

sin^-1 (-1/2)+tan^-1 (sqrt(3))= (A) -pi/6 (B) pi/3 (C) pi/6 (D) none of these

Check whether, the following equations can represent a progressive (travelling wave) (a) y=A cos(x^(3)-vt) (b) x=Ae^((vt-y))(c) y=A log((x)/(v)-t)

If sin^(-1)(tan(pi/4))-sin^(-1)(sqrt(3/y))-(pi)/6=0 and x^(2)=y then x is equal to

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)

If x=3tant and y=3sect then find (d^2y)/(dx^2) at x=(pi)/(4) (a) 3 (b) (1)/(6sqrt2) (c) 1 (d) (1)/(6)