Motion (1 Viewer)

[Lukybear]

Solve the differential equation a (d2x/dt2) + 16x = 0 (SHM)

Also, when solving strait line motion equations. How do we take the signs. For example

A particle moves in a straight line so that at any time t, its displacement from a fixed origin is x and its velocity is v. If acceleration is 3x^2 and v=-sqrt2 x=1 when t=0, find x as a function of t.

I.e. when finding
v^2 = 2x^3
how do we take the sign?
is it v=sqrt 2x^3 or v=-sqrt2x^3

 

[ninetypercent]

v = -sqrt 2 when x = 1
so take the negative sign
i.e. v = -sqrt 2x^3
when x = 1, v = -sqrt 2

on the other hand, if you took the positive square root
v = sqrt 2x^3
when x = 1, v = sqrt 2
this would not satisfy the condition give in the question that when x = 1, v = -sqrt 2

 

[Trebla]

Solve the differential equation a (d2x/dt2) + 16x = 0 (SHM)

You can just quote the general solution of the displacement equation of SHM or do it analytically (which is not required by the syllabus)

Let v = dx/dt
Note that: d²x/dt² = d(v²/2)/dx
=> d(v²/2)/dx = - 16x
v² = - 16x² + c₁
v = √ (c₁ - 16x²) if v is non-negative
v = - √ (c₁ - 16x²) if v is negative

From positive case:
dx/dt = √ (c₁ - 16x²)
dt/dx = 1/√ (c₁ - 16x²)
= (1/4) (1/√ (c₁/16 - x²))
t = - (1/4)cos^-1(4x/√c₁) + c₂
x = (√c₁/4) cos(4c₂ - 4t)
= (√c₁/4) cos(4t - 4c₂) as the cosine function is even
= a cos (4t + α)

(Can easily check the negative case leads to same result)