91Ó°ÊÓ

In the SHM, the energy is conserved and the total mechanical energy E has an expression in terms of the force constant k and amplitude A as given,

$\begin{array}{r}E=K+U\\ \frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2\end{array}$

Where, K is the kinetic energy and U is the potential energy, also, A is the amplitude, k is the force constant, m is the mass, v is the velocity or speed of the oscillation and x is the displacement.

Given, force constant (k) = 155 N/m, mass (m) = 0.175 kg, velocity (v) =0.815 m/s, and displacement (x) = 0.03 m.

Now, solve the equation (1) for A,

$\begin{array}{r}\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2\\ k{A}^2=m{v}^2+k{x}^2\\ A=\sqrt{\frac{m{v}^2+k{x}^2}{k}}\\ A=\sqrt{\frac{\left(0.175\mathrm{kg}\right)(0.815\mathrm{m}/\mathrm{s}{)}^2+(155\mathrm{N}/\mathrm{m}\left)\right(0.03\mathrm{m}{)}^2}{155\mathrm{N}/\mathrm{m}}}\\ =4.06×{10}^{−2}\mathrm{m}\\ =4.06\mathrm{cm}\end{array}$

Hence, the amplitude of the motion is A = 4.06 cm.

The glider reaches the maximum velocity when the potential energy, U is zero, where all energy converts to kinetic energy, K.

So, from equation (1), expression for the maximum velocity, you get

$\begin{array}{r}E=K+U\\ \frac{1}{2}k{A}^2=\frac{1}{2}m{v_{\text{max}}}^2+0\\ v_{max}=\sqrt{\frac{k{A}^2}{m}}\\ v_{max}=\sqrt{\frac{(155\mathrm{N}/\mathrm{m}){\left(4.06×{10}^{−2}m\right)}^2}{0.175\mathrm{kg}}}\\ =1.21\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the maximum speed of the glider is ${\mathit{v}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{21}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$.

Use the expression $a_{max}$and$\mathrm{Ó\neg }$ you get the relationship between the frequency, F and the spring constant, k from Newton's law of motion.

Now,

$\begin{array}{r}F=m{a}_{max}\\ k{x}_{max}=m\left({\mathrm{Ó\neg }}^2{x}_{max}\right)\\ \begin{array}{rl}\mathrm{Ó\neg }& =\sqrt{\frac{k}{m}}\\ \mathrm{Ó\neg }& =\sqrt{\frac{155N/m}{0.175\mathrm{kg}}}\\ & =29.8\mathrm{rad}/\mathrm{s}\end{array}\end{array}$

Hence, the angular frequency of the oscillations is $\mathit{Ó\neg }\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{8}\mathbf{}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}\mathit{s}$.