91Ó°ÊÓ

1. Spring constant of the spring, k= 400 N/m
2. Position at time t, x =0.100 m
3. Velocity at time t, v = - 13.6 m
4. Acceleration at time t, a = 123 m/s²

Using the formula of acceleration of SHM and using the relation between angular velocity and frequency, we can find the frequency of oscillation.

Using the formula of angular velocity in terms of spring constant and mass, we can find the mass of the block. Then, using the law of energy conservation, we can find the amplitude of the motion.

Formula:

The acceleration of a body in oscillation,

$a=-{\mathrm{Ó\neg }}^{2}x$ (i)

The angular frequency of a body in oscillation,

$\mathrm{Ó\neg }=2\mathrm{Ï€}f$ (ii)

The angular frequency of a spring,

$\mathrm{Ó\neg }=\sqrt{k/m}$ (iii)

The potential energy of a body in oscillation,

$U=\frac{1}{2}k{x}_m^2$ (iv)

The kinetic energy of a body in motion,

$K=\frac{1}{2}m{v}^2$ (v)

Using equation (i) and the given values, we get the angular frequency as:

$$\begin{gathered} \mathrm{Ó\neg }=\sqrt{-\frac{a}{x}}(âˆ\mu a=-123{\text{m/s}}^{\text{2}}\&x=0.100\text{m}) \\ =\sqrt{-\frac{\left(-123\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)}{0.100\text{m}}} \\ =35.07\text{rad/s} \end{gathered}$$

Again using the equation (ii), we get the frequency as:

$$\begin{gathered} f=\frac{\mathrm{Ó\neg }}{2\mathrm{Ï€}} \\ =\frac{35.07\frac{\text{rad}}{\text{s}}}{2\mathrm{Ï€}} \\ =5.58\text{Hz} \end{gathered}$$

Therefore, thefrequency of oscillation is 5.58 Hz

Using equation (iii), we get the mass as:

$$\begin{gathered} m=\frac{k}{{\mathrm{Ó\neg }}^2}(âˆ\mu k=400\text{N/m}\&\mathrm{Ó\neg }=35.07\text{rad/s}) \\ =\frac{400\text{N/m}}{{\left(35.07\frac{\text{rad}}{\text{s}}\right)}^2} \\ =0.325\text{kg} \end{gathered}$$

Therefore, the mass of the block is 0.325 Kg

Using equation (iv) & (v), the total energy of the system can be given as:

$\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

Therefore, according to energy conservation law,

$$\begin{gathered} \frac{1}{2}k{x}_m^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2 \\ {x}_m^2=x^2+\frac{m}{k}{v}^2 \\ {x}_m=\sqrt{x^2+\frac{m}{k}{v}^2} \\ =\sqrt{{\left(0.100m\right)}^2+\frac{\left(0.325\text{kg}\right)}{\left(400\text{N/m}\right)}{\left(-13.6\frac{\text{m}}{\text{s}}\right)}^2} \\ =0.400\text{m} \end{gathered}$$

Therefore, the amplitude of the motion is 0.400 m.