Question

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at (e) the

Answer 1

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
$E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
$E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
$E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
$\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
$v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b)  the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part. 
The total mechanical energy is:
$E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since 
$E=U+K$
we find the kinetic energy when the glider is at this position:
$K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
$K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
$a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find 
$a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
$E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$