91Ó°ÊÓ

• The mass of the block, m=0.1kg.
• The angular frequency of the oscillations,$\mathrm{Ó\neg }=10\mathrm{rad}/\mathrm{s}$.
• The amplitude of the oscillations,$x_m=10\mathrm{cm}\mathrm{or}0.1\mathrm{m}$.
• The phase constant, $\mathrm{Ï•}=\frac{\mathrm{Ï€}}{2}$.

Using the expression of SHM for maximum velocity and maximum acceleration, we can find their respective values and at which position they occur.

Using the expression for angular frequency and force we can find frequency and force.

Formula:

The frequency of the oscillations, $f=\mathrm{Ó\neg }/2\mathrm{Ï„}\mathrm{Ï„}$ (i)

The maximum velocity of the oscillation,$v_{max}=\mathrm{Ó\neg }{x}_m$ (ii)

The magnitude of the maximum acceleration,$\left|a_{max}\right|={\mathrm{Ó\neg }}^2{x}_m$ (iii)

The force of spring as per Hook’s law, $F=kx=m{\mathrm{Ó\neg }}^2{x}_m$ (iv)

Using equation (i), the value of oscillation frequency is given as:

$\begin{array}{l}\mathrm{f}=\frac{10\mathrm{rad}/\mathrm{s}}{2×3.14}\\ =1.592\mathrm{Hz}\\ ≈1.6\mathrm{Hz}\end{array}$

Hence the value of frequency is 1.6 Hz.

Using equation (ii), the maximum speed is given as:

$$\begin{gathered} v_{max}=10rad/s×0.1\mathrm{m} \\ =1.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of maximum speed is 1.0 m/s.

$X\left(v_{max}\right)$: The maximum speed occurs at t=0.

$X\left(v_{max}\right)=10\cos \left[\frac{\mathrm{Ï„}\mathrm{Ï„}}{2}\right]$

Hence, the required position occurs at 0m.

Using equation (iii), the magnitude of maximum acceleration is given as:

$$\begin{gathered} \left|a_{max}\right|={\left(10\mathrm{rad}/\mathrm{s}\right)}^2×0.1\mathrm{m} \\ =10\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of maximum acceleration is $10\mathrm{m}/{\mathrm{s}}^2$.

$X\left(a_{max}\right)$: The magnitude of acceleration is greatest at extreme positions.

$$\begin{gathered} X\left(a_{max}\right)=x_m \\ =0.1\mathrm{m} \end{gathered}$$

Hence, the required position is 0.1 m.

We know that the spring constant is given as:

$$\begin{gathered} \mathrm{k}={\mathrm{mÓ\neg }}^2 \\ =10\mathrm{kg}×110\mathrm{rads}{)}^2 \\ =10\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, as per equation (iv), we have the force on the block as:

$F=-10×\mathrm{N}$

Hence, the value of the force is -10xN.