A160 g block attached to a spring with spring constant 2.6 n/m oscillates horizontally on a frictionless table. its velocity is 21 cm/s when x0 = -4.9 cm . a)what is the amplitude of oscillation? b)what is the block's maximum acceleration? c)what is the block's position when the acceleration is maximum? d)what is the speed of the block when x1 = 3.3 cm ?

A) 0.071 m (7.1 cm)

The table is frictionless, so the mechanical energy of the system is conserved, and it is the sum of the kinetic energy and the potential energy of the spring:

Where

m = 160 g = 0.16 kg is the mass of the block

v = 21 cm/s = 0.21 m/s is the velocity of the block

k = 2.6 N/m is the spring constant

x = -4.9 cm = -0.049 m is the displacement of the block with respect to the equilibrium position

Substituting into the formula, we find:

This is the mechanical energy, and it is constant at any moment of the motion. We also known that the amplitude of the oscillation corresponds to the maximum displacement , and when the system is at maximum displacement, the velocity is zero, and all the energy is elastic potential energy:

From which we find the amplitude:

B) 1.15 m/s^2

The angular frequency of the system is given by

And the maximum acceleration is given by:

C) x = 0.071 m (7.1 cm)

In a simple harmonic motion, the acceleration of the block is related to the displacement x(t) by the formula

Therefore, we see that the acceleration is proportional to the displacement: this means that when the acceleration is maximum, the position x(t) is also maximum, so the block is located at maximum displacement, which corresponds to the amplitude:

D) 0.25 m/s (25 cm/s)

Again, we can solve the problem by using the conservation of energy:

where this time we know:

E = 0.0066 J

m = 0.16 kg

v = ?

k = 2.6 N/m

x = 3.3 cm = 0.033 m

Substituting, we find

option number c is the right answer.