How can I calculate maximum velocity in simple harmonic motion?

Question

How can I calculate maximum velocity in simple harmonic motion?

I know that the equation for velocity is

$V = - ω A sin (ω t + ϕ$ $V= -omega Asin(omega t +phi$ )

supposing $ϕ$ $phi$ to be zero , cuz if the object is released from the mean position then, at the mean position displacement is zero so,

$sin ϕ$ $sin phi$ = 0 → $ϕ$ $phi$ = 0.

I know from the velocity time graph for SHM that max velocity = $A ω$ $Aomega$ .

That is the argument must be -1

How will that be possible ? the argument can only be zero for $2 \frac{π}{3}$ $2pi/3$ right?

Am I missing something?

1 Answer

Impact of this question

22062 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-26T14:30:44

$Please see the following explanations.$ $"Please see the following explanations."$

Explanation:

enter image source here

we recommend that you first watch the animation carefully
Let P be a point rotating around the O point at a constant linear velocity(u).
The projection of P on the x-axis is point B.
The movement of point B is limited between A and F.
The simple harmonic motion is the action of point B.
Please note that the velocity vector changes direction.
At points A and F, the velocity of B is zero(green vector).
The greatest velocity B has is at O. $θ = \frac{π}{2} and θ = \frac{3 π}{2}$ $theta=pi/2 and theta=(3pi)/2$

$x = r cos θ$ $x=r cos theta$

$θ = ω . t$ $theta=omega.t$

$x = r . cos (ω t)$ $x=r. cos(omega t)$

$\frac{d x}{d t} = v$ $(d x)/(d t)=v$

$\frac{d x}{d t} = v = - r ω sin ω t$ $(d x)/(d t)=v=-r omega sin omega t$

$if ω t = \frac{π}{2} or ω t = \frac{3 π}{2}, v = - r ω (maximum velocity)$ $"if "omega t=pi/2" or "omega t=(3pi)/2" , "v=-r omega("maximum velocity")$

$v_{max} = - r ω$ $v_("max")=- r omega$

$or :$ $"or :"$

$y = r sin θ$ $y=r sin theta$

$\frac{d x}{d t} = v = - ω y$ $(d x)/(d t)=v=- omega y$

$x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

$y = \sqrt{r^{2} - x^{2}}$ $y=sqrt(r^2-x^2)$

$v = - ω \sqrt{r^{2} - x^{2}}$ $v=- omega sqrt(r^2-x^2)$

$if x = 0 v = v_{max}$ $"if "x=0 " "v=v_("max")$

Science

Math

Humanities

... and beyond

How can I calculate maximum velocity in simple harmonic motion?

1 Answer

Explanation:

Related questions

Impact of this question