Question

In Concept Simulation 10.3 you can explore the concepts that are important in this problem. A block of mass m = 0.563 kg is fastened to an unstrained horizontal spring whose spring constant is k= 94.6 N/m. The block is given a displacement of +0.106 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (with sign) that the spring exerts on the block just before the block is released? (b) Find the angular frequency  of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d)Determine the magnitude of the maximum acceleration of the block.

Answer 1

See the diagram :

at X = 0, the spring is unstretched or unstrained.

suppose the body is displaced by x = xo, and released.

As the body moves from unstretched position, spring force (Fs) increases, and it is given by Fs = - kx,

where k = spring constant and x is displacement from unstretched postion.

also, -ve sign indicates that this force always direct towards the unstretched position.

Given :

mass of block = m = 0.563 kg

spring constant = k = 94.6 N/m

xo = 0.106 m.

a) Just before the block is released, it is at x = xo, Spring Force on it is Fs = - kxo (This magnitude is maximum).

=> Fs = - 10.03 N

b) At any value of x, the force on the block is Fs = -kx, and this force produces acceleration (say a).

therefore, ma = -kx

=> a/x = -k/m

|a/x| = |k/m|

angular frequncy of this oscillation is = (acceleration per unit displacment )

=> .

c) speed of block at any x is given by:

for maximum speed, x = 0, so vmax  = xo = 12.96 x 0.106 = 1.37 m/s.

d) When the block is at x = xo, the force on it is maximum and so it's acceleration is also maximum.

Therefore, amax = Fmax/m = kxo/m = 17.81 m/s2.