A particles starts from point x=-sqrt(3)...

A particles starts from point `x=-sqrt(3)/(2) A` and move towards negative extrene as shown . If the time period of oscillation is T , them

The equation of the SHM is `x= A sin ((2pi)/(T)+(4pi)/(3))`

The time taken by the particle tto go directly from its initial position to negative extreme is `(T)/(12)`

The time taken by the particle to reach at mean position is `(T)/(3)`

The equation of the SHM is ` x= A sin ((2pi)/(T)+(pi)/(3))`

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The correct Answer is:

A, B, C

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A particle moves such that acceleration is given by a= -4x . The period of oscillation is

`pi`

`(2)/(pi)`

`(1)/(pi)`

`2pi`

`X=A sin ((2pi)/(T)t+(pi)/(6))`

`X=A sin((2pi)/(T)t+(5pi)/(6))`

`X=A cos((2pi)/(T)t+(pi)/(6))`

`X=A cos ((2pi)/(T)t+(pi)/(3))`

`x = A sin ((2pi)/(T)t + (pi)/(6))`

`x = A sin((2pi)/(T)t + (5pi)/(6))`

`x = A cos ((2pi)/(T)t + (pi)/(6))`

`x = Acos ((2pi)/(T)t + (pi)/(3))`