iTeachChem•9mo ago

Oscillations Doubt

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

PercyOP•3/26/25, 9:16 AM

@Nimboi

SirLancelotDuLac•3/26/25, 10:07 AM

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

TeXit•3/26/25, 10:07 AM

SirLancelotDuLac

SirLancelotDuLac•3/26/25, 10:08 AM

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

Nimboi•3/26/25, 11:16 AM

@Gyro Gearloose

prasadam123•3/26/25, 11:20 AM

@Percy what is the answer

Pprasadam123 @Percy what is the answer

PercyOP•3/26/25, 11:32 AM

option d

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

PercyOP•3/26/25, 11:34 AM

Are you able to understand this explanation

PPercy Are you able to understand this explanation

SirLancelotDuLac•3/26/25, 11:36 AM

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.

SSirLancelotDuLac Well, we can get $v=x_{0}sin(t)$, so $v^{2}={x_{0}}^{2}(sin^{2})=4{x_{0}}^{2}(si...

SirLancelotDuLac•3/26/25, 11:37 AM

This is another way for this question.