Question

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.

Answer 1

Given:
Equation of motion of the particle executing S.H.M.,
$$\begin{gathered} x=\left(2.0\mathrm{cm}\right)\sin \left[\left(100s^{-1}\right)t+\frac{\mathrm{\pi }}{6}\right] \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{particle},m=10\mathrm{g} \end{gathered}$$ ...(1)
General equation of the particle is given by,
$x=A\sin (\omega t+\varphi )$ ...(2)

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

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

$$\begin{gathered} \mathrm{Time}\mathrm{period}\mathrm{is}\mathrm{calculated}\mathrm{as}, \\ T=\frac{2\mathrm{\pi }}{\omega }=\frac{2\mathrm{\pi }}{100}=\frac{\mathrm{\pi }}{50}\mathrm{s} \\ =0.063\mathrm{s} \end{gathered}$$

Also, we know -
$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{m}{k}} \\ \mathrm{where}\mathrm{k}\mathrm{is}\mathrm{the}\mathrm{sprin}g\mathrm{constant}. \\ \Rightarrow T^2=4{\mathrm{\pi }}^2\frac{m}{k} \\ \Rightarrow k=\frac{4{\mathrm{\pi }}^{2}m}{T^2}={10}^5\mathrm{dyne}/\mathrm{cm} \\ =100\mathrm{N}/\mathrm{m} \end{gathered}$$


(b) At t = 0 and x = 2 cm $\sin \frac{\mathrm{\pi }}{6}$
$=2\times \frac{1}{2}=1\mathrm{cm}\mathrm{from}\mathrm{the}\mathrm{mean}\mathrm{position},$

We know:
x = A sin (ωt + ϕ)

Using $v=\frac{dx}{dt},$ we get:
v = Aω cos (ωt + ϕ)
$$\begin{gathered} =2\times 100\cos \left(0+\frac{\mathrm{\pi }}{6}\right) \\ =200\times \frac{\sqrt{3}}{2} \\ =100\sqrt{3}{\mathrm{cms}}^{-1} \\ =1.73{\mathrm{ms}}^{-1} \end{gathered}$$


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