Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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A Particle Having Mass 10 G Oscillates According to the Equation X = (2.0 Cm) Sin [(100 S−1)T + π/6]. - Physics

Sum

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.

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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,

\[x = A\sin(\omega t + \phi)\] ...(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

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 12 Simple Harmonics Motion
Q 5 | Page 252
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