91Ó°ÊÓ

1. Amplitude of watch,${\mathrm{θ}}_{\mathrm{m}}=\mathrm{π}\mathrm{rad}$.
2. Time period,$T=0.500\sec$.

Using the given information and the standard equation of SHM, we can find the required values.

Formula:

Angular frequency of a body, $\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$….. (i)

Angular displacement of a body,$\mathrm{Î\copyright }\left(t\right)={\mathrm{θ}}_m\cos \left(\mathrm{Ó\neg }t\right)$…â¶Ä¦(¾±¾±)

Angular velocity of a body,$\mathrm{Î\copyright }=-\left(\frac{2\mathrm{Ï€}}{T}\right){\mathrm{θ}}_m\sin \left(\frac{2\mathrm{Ï€}t}{T}\right)$…â¶Ä¦â¶Ä¦.(¾±¾±¾±)

Angular acceleration of a body,$\mathrm{Î\pm }=-\mathrm{Î\copyright }{\left(\frac{2\mathrm{Ï€}}{T}\right)}^2{\mathrm{θ}}_m\cos \left(\frac{2\mathrm{Ï€}t}{T}\right)$…â¶Ä¦â¶Ä¦(¾±±¹)

The maximum angular speed of a body, ${\mathrm{Î\copyright }}_m=\frac{2\mathrm{Ï€}{\mathrm{θ}}_m}{T}$…â¶Ä¦..(±¹)

Using equation (v) and the given values, we get the maximum angular speed as:

$$\begin{gathered} {\mathrm{Î\copyright }}_m=\frac{2\mathrm{Ï„}\mathrm{Ï„}\left(\mathrm{Ï„}\mathrm{Ï„}\right)}{0.50} \\ =39.5\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the angular speed value is found to be$39.5rad/s$.

Using equation (i) and substituting the values, we get the angular speed at required displacement as:

$$\begin{gathered} \mathrm{Î\copyright }\left(t\right)=\mathrm{Ï€}\cos \left(\mathrm{Ó\neg }t\right) \\ \frac{\mathrm{Ï€}}{2}=\mathrm{Ï€}\cos \left(\mathrm{Ó\neg }t\right) \\ \cos \left(\mathrm{Ó\neg }t\right)=\frac{1}{2} \\ \left(\mathrm{Ó\neg }t\right)=\frac{\mathrm{Ï€}}{3} \end{gathered}$$

Calculate the value of angular velocity as below.

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T} \\ =\frac{2\left(3.14\right)}{0.50} \\ =12.56\mathrm{rad}/\mathrm{s} \end{gathered}$$

Now, the angular velocity using equation (iii) and the given values is given as:

$$\begin{gathered} \mathrm{Î\copyright }=-\left(\mathrm{Ó\neg Ï„Ï„}\right)\sin \left(\mathrm{Ó\neg t}\right) \\ =-\left(12.56\right)\left(3.14\right)\left(\frac{\sqrt{3}}{2}\right) \\ =-34.11\mathrm{rad}/\mathrm{s} \\ ≈-34.11\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the angular speed value is found to be .

Using equation (iii) and the given values, the angular acceleration is given as:

$$\begin{gathered} \mathrm{Î\pm }\left(t\right)=-{\left(\mathrm{Ó\neg }\right)}^2\mathrm{θ} \\ =-{\left(12.56\right)}^2\left(\frac{\mathrm{Ï„}\mathrm{Ï„}}{4}\right) \\ =-123.8\mathrm{rad}/{\mathrm{s}}^2 \\ ≈-124\mathrm{rad}/{\mathrm{s}}^2 \\ \left|\mathrm{Î\pm }\left(\mathrm{t}\right)\right|=124\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of angular acceleration is found to be .