91Ó°ÊÓ

Frequency is defined as the number of oscillations done by an object per second.

The amplitude of motion can be defined as the maximum distance that an object travels before coming back to its original position.

The period can be defined as the time taken required to complete one cycle or oscillation.

Angular frequency is a quantity that measures how quickly an object oscillates with respect to time.

(a) Calculate the displacement using the formula

In a simple harmonic motion, the displacement is given by

$x=A\cos \left(\mathrm{Ï–}t-\mathrm{Ï•}\right)$

Consider the given data as below.

When time, $t=0$s and the displacement is $x=A$.

$$\begin{gathered} A=A\cos \mathrm{Ï•} \\ \mathrm{Ï•}={\cos }^{-1}\left(1\right) \\ =0rad \end{gathered}$$

The angular frequency is given by

$$\begin{gathered} \mathrm{Ï–}=\frac{2\mathrm{Ï€}}{T} \\ =\frac{2\mathrm{Ï€}}{1.5} \\ =4.2rad/s \end{gathered}$$

The displacement at $t=0.5s$ will be

$$\begin{gathered} x=0.24\cos \left(4.2×0.5-0\right) \\ =0.24\cos \left(2.1\right) \\ =0.121m \end{gathered}$$

Hence, the displacement is$x=-0.121m$.

In an SHM, the force is given by

$F=-kx$

The angular frequency is given by

$\mathrm{Ï–}=\sqrt{\frac{k}{m}}$

The equation of force will become $F=-M{\mathrm{Ï–}}^{2}x$.

$$\begin{gathered} F=-{\left(4.2\right)}^2×0.02×\left(-0.121\right) \\ =0.043N \end{gathered}$$

Hence, the force is $F=0.043N$ and the force is acting in the positive x-axis.

The minimum time required for the bolt to move from its initial position $A=0.24m$ to the point $x=-0.18m$will be

$$\begin{gathered} -0.18=0.24\cos \left(4.2t-0\right) \\ t=\frac{1}{4.2}{\cos }^{-1}\left(\frac{-0.18}{0.24}\right) \\ t=0.58s \end{gathered}$$

Hence, the time is$0.58s$.

In an SHM, the velocity is given by

$$\begin{gathered} V_x=\underset{-}{+}\sqrt{\frac{k}{m}}\sqrt{A^2-x^2} \\ {V}_x=\underset{-}{+}\mathrm{Ï–}\sqrt{A^2-x^2} \\ =\underset{-}{+}4.2\sqrt{0.{24}^2-{\left(-0.18\right)}^2} \\ =\underset{-}{+}0.67m/s \end{gathered}$$

Hence, the velocity is $V_x=0.67m/s$.