Question

A transverse wave is travelling along a string from left to right. The figure represents the shape of the string (snap-shot) at a given instant. At this instant how many points will have downward magnitude of velocity.

Answer 1

Equation of travelling wave is given by:
$y=Asin(kx-\omega t)$
Differentiating with respect to time.
$v_y=-A\omega cos(kx-\omega t)$
The snapshot is taken at an instant ($t$ is fixed), without any loss of generality let us take $t=0$, Then
$v_y=-A\omega cos\left(kx\right)$
For $v_y$ to be negative, $cos$ term must be positive and $x$ must lie in the range $(0,\frac{\pi }{2})$ and[$(\frac{3\pi }{2},2\pi )$
In other words $kx$ must lie outside of $\frac{\lambda }{4}and\frac{3\lambda }{4}$
Hence, $1,2,8$ satisfy this.