Answer 1

Given:
Equation of motion of the particle executing S.H.M , 

\[x = \left( 2 . 0 cm \right) \sin \left[ \left( 100 s^{- 1} \right)t + \frac{\pi}{6} \right]\]

\[\text { Mass of the particle} , m = 10 g\]...(1)

General equation of the particle is given by,

\[x = A\sin(\omega t + \phi)\] ...(2)

On comparing the equations (1) and (2) we get:

(a) Amplitude, A is 2 cm.
   Angular frequency, ω is 100 s⁻¹​.

\[\text { Time period is calculated as }, \]

\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{100} = \frac{\pi}{50}s\]

\[ = 0 . 063 s\]

Also, we know -

\[T = 2\pi\sqrt{\frac{m}{k}}\]

\[\text { where k is the spring constant }. \]

\[ \Rightarrow T^2 = 4 \pi^2 \frac{m}{k}\]

\[ \Rightarrow k = \frac{4 \pi^2 m}{T^2} = {10}^5 dyne/cm\]

\[ = 100 N/m\]

(b) At t = 0 and x = 2 cm

\[\sin\frac{\pi}{6}\] \[= 2 \times \frac{1}{2} = 1 \text {cm} \text { from the mean position },\]

We know:     
     x = A sin (ωt + ϕ)

Using

\[v = \frac{\text {dx}}{\text { dt}},\]  we get:

 v = Aω cos (ωt + ϕ)

\[= 2 \times 100 \cos \left( 0 + \frac{\pi}{6} \right)\]

\[ = 200 \times \frac{\sqrt{3}}{2}\]

\[ = 100\sqrt{3} {\text { cms }}^{- 1} \]

\[ = 1 . 73 {\text {ms}}^{- 1}\]

(c) Acceleration of the particle is given by,
    a = \[- \omega^2\]x
          = 100²×1 = 10000 cm/s²