Oscillations Doubt

[buoyant orchid]

How to identify the mean position and extreme position in these type of questions?

[buoyant orchid]

@pastel gazelle

[pale jacinth]

Well, we can get $v=x_{0}sin(t)$, so $v^{2}={x_{0}}^{2}(sin^{2})=4{x_{0}}^{2}(sin^{2}(t/2)(1-sin^{2}(t/2)))=4x_{0}(1-x_{0})$

[hazy flare]

SirLancelotDuLac

[pale jacinth]

So, we can just look at the quadratic obtained and mark off d as the answer ig.

[pastel gazelle]

<@&1227987967939838003>

[tight root]

@buoyant orchid what is the answer

[buoyant orchid]

option d

[buoyant orchid]

Are you able to understand this explanation

[pale jacinth]

Yep. The mean position of x=a+bsin(theta) is given by a right? So basically the solution says that x/2 is the mean position and hence velocity at that point will be maximum and at x=0 and x_0 the extreme positions are obtained and hence velocity at those points are zero.

[pale jacinth]

This is another way for this question.☝️

[tight root]

by differentiating twice you get the accn eqn and equating it to zero will give you the time at which the particle is at mean position

[buoyant orchid]

Thank you so much

+solved @pale jacinth @tight root