Which statement describes the velocity of a mass oscillating on a spring? the velocity is constant throughout the motion. the velocity remains the same throughout the motion. the velocity is zero at the maximum displacement. the velocity is greatest at the maximum position. the velocity is least at the equilibrium position.

The velocity is zero at the maximum displacement.

Explanation:

As we know that in oscillations on a spring the force due to spring is always towards the mean position.

So here when an object is connected with a spring the force is given by

F = - kx

now here as object will move away from the mean position then it's velocity will continuously decreases and it comes to rest at its maximum displacement position.

Then after that its spring force will pull it towards the mean position and the mass will accelerate to come at its mean position. So here the velocity of that mass will be maximum when it will reach to its mean position.

So here correct answer will be

The velocity is zero at the maximum displacement.

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

seeing the graph would