A 0.500-kg glider, attached to the end of an ideal spring with force constant k = 450 N/m, 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 x = -0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at x = -0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

Formula used,

$\mathbf{E}\mathbf{=}{\mathbf{U}}_{\mathbf{s}}\mathbf{+}\mathit{K}$

a)

The maximum speed of the block will be at x=0, which means when the block reaches maximum speed its potential energy is 0.

$\begin{array}{r}\mathrm{E}=\mathrm{K}+0\\ \frac{1}{2}{\mathrm{KA}}^2=\frac{1}{2}{\mathrm{mv}}^2\\ {\mathrm{KA}}^2={\mathrm{mv}}_{max}^2\\ {\mathrm{v}}_{max}^2=\frac{{\mathrm{KA}}^2}{\mathrm{m}}\\ {\mathrm{v}}_{max}=\sqrt{\frac{{\mathrm{KA}}^2}{\mathrm{m}}}\\ {\mathrm{v}}_{max}=A\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\\ {\mathrm{v}}_{max}=0.040\times \sqrt{\frac{450}{0.50}}\\ {\mathrm{v}}_{max}=1.20\mathrm{m}/\mathrm{s}\end{array}$

We know that,

$\mathbf{E}\mathbf{=}{\mathbf{U}}_{\mathbf{s}}\mathbf{+}\mathit{K}$

Hence,

b)

$\begin{array}{r}\frac{1}{2}K{A}^2=\frac{1}{2}K{x}^2+\frac{1}{2}m{v}^2\\ K{A}^2=K{x}^2+m{v}^2\\ v^2=\frac{K\left(A^2-x^2\right)}{m}\\ v=\pm \sqrt{\frac{K\left(A^2-x^2\right)}{m}}\\ v=\pm \sqrt{\frac{450\times \left((0.040{)}^2-(-0.015{)}^2\right)}{0.50}}=\pm 1.11\mathrm{m}/\mathrm{s}\end{array}$

c)

We know that maximum acceleration is x=A when speed is equal to 0.

$\begin{array}{l}-\mathrm{kx}={\mathrm{ma}}_{\mathrm{x}}\\ {\mathrm{a}}_{\mathrm{x}}=\frac{-\mathrm{Kx}}{\mathrm{m}}\end{array}$

Maximum acceleration will be x= A

$\begin{array}{r}\left|a_{x,max}\right|=\frac{|-\mathrm{KA}|}{m}\\ \left|a_{x,max}\right|=\frac{|-\mathrm{KA}|}{\mathrm{m}}\\ {\mathrm{a}}_{\mathrm{x},max}=\frac{450\times 0.040}{0.50}=36\mathrm{m}/{\mathrm{s}}^2\end{array}$

d)

We know that,

$\begin{array}{r}{a}_x=\frac{-Kx}{m}\\ a_x=\frac{-450\times -0.015}{0.50}=+13.5\mathrm{m}/{\mathrm{s}}^2\end{array}$

e)

The total mechanical energy of the glider at any point in its motion is,

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{K}{\mathit{A}}^{\mathbf{2}}$

$\begin{array}{l}E=\frac{1}{2}\times 450\times {\left(0.040\right)}^2\\ E=036J\end{array}$

Hence,

1. The maximum speed of the glider is 1.20 m/s.
2. The speed of the glider when x= -0.015 m is 1.11 m/s.
3. The magnitude of the maximum acceleration of glider is 36 $m/s^2$.
4. The acceleration of the glider at x=-0.015 m is +13.5 $m/s^2$.
5. The total mechanical energy of the glider at any point in its motion is 0.36 J.