Max and Min velocity of SHM

Let equation of shm be

X=a sin (wt) ---- (1) [w->omega or angular velocity] [x-> displacement of particle from mean position]

Also from 1 --- sin(wt) = x/a

Differentiating wrt time dX/dt = v = aw cos(wt)

Cos (y)= (1-sin^2(y))^1/2

So velocity(v) = aw (1-(x/a)^2)^1/2

Simplifying

V= w (a^2- x^2)^1/2

Clearly under root will never give -ve value So max is when x=o (mean position) ie aw And min is when x=a (extreme position) ie 0

Vmax + Vmin = aw = 2 * 5 = 10

$y=A sin(\omega t)$

$v=\frac{dy}{dt}=A \omega cos(\omega t)$

$v_{\max}= A \omega$

since $v_{\min}=0$ ,

$v_{total}=v_{\max}=5 \times 2 =\boxed{10}$

Easiest damn easy soty.
Ttttt

Minimum Velocity is $0$ at extreme postion or at $y=A$ where $A$ is the Amplitude.

Maximum Velocity is $A\omega = 10 \frac{m}{s}$ at mean postion or at $y=0$ .

Therefore the sum is $10 \frac{m}{s}$