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Simple Harmonic Motion — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsSimple Harmonic Motion5 Marks
Question
A particle having mass 10g oscillates according to the equation $\text{x}=(2.0\text{cm})\sin\big[(100\text{s}^{-1})\text{t}+\frac{\pi}{6}\big].$ Find (a) the amplitude, the time period and the spring constant (b) the position, the velocity and the acceleration at t = 0.
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Answer

$\text{x}=(2.0\text{cm})\sin\Big[(100\text{s}^{-1})\text{t}+\Big(\frac{\pi}{6}\Big)\Big]$m = 10g.
  1. Amplitude = 2cm
$\omega=100\sec^{-1}$
$\therefore\text{T}=\frac{2\pi}{100}=\frac{\pi}{50}\sec=0.063\sec.$
We know that $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
$\Rightarrow\text{T}^2=4\pi^2\times\frac{\text{m}}{\text{k}}$
$\Rightarrow\text{k}=\frac{4\pi^2}{\text{T}^2}\text{m}$
$=10^5\text{dyne/cm}=100\text{N/m}$ $\Big[\text{because}\ \omega=\frac{2\pi}{\text{T}}=100\sec^{-1}\Big]$
  1. At t = 0
$\text{x}=2\text{cm}\sin\Big(\frac{\pi}{6}\Big)=2\times\Big(\frac{1}{2}\Big)=1\text{cm}.$ from the mean position.
We know that $\text{x}=\text{A}\sin(\omega\text{t}+\phi)$
$\text{v}=\text{A}\cos(\omega\text{t}+\phi)$
$=2\times100\cos\big(0+\frac{\pi}{6}\big)=200\times\frac{\sqrt{3}}{2}$
$=100\sqrt{3}\sec^{-1}=1.73\text{m/s}$
  1. $\text{a}=-\omega^2\text{x}=100^2\times1$
$=100\text{m/s}^2$

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