A particle having mass 10 g oscillates according to the equation *x* = (2.0 cm) sin [(100 s^{−1})*t* + π/6]. Find (a) the amplitude, the time period and the spring constant. (c) the position, the velocity and the acceleration at *t* = 0.

#### Solution

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,

(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

*x*= A sin (ω

*t*+ ϕ)

Using

*v*=

*A*ω cos (ω

*t*+ 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*

= 100^{2}×1 = 10000 cm/s^{2}