Question
If a mass of 0.8 kg is moving in simple harmonic motion starting from equilibrium position. The dimensions of the body of mass 1.0 m and if the period is 11/7 sec then 0.6 m. Find velocity of the particle at displacement and also write the equation of motion.

Answer

Given :
$\begin{array}{l}m=0.8 kg \\A=1.0 m\end{array}$
Period $\quad T=\frac{11}{7} sec \therefore$ Frequency $(n)=\frac{7}{11}$ per sec.
$ \begin{aligned}\text {Displacement}\quad(x) & =0.6 m \\v & =?\end{aligned}$
Equation of the motion $=$ ?
between velocity (v) and displacement ( x ) in simple harmonic motion, their relation :
$\begin{aligned}v & =\omega \sqrt{A^2-x^2}\ldots\ldots (1) \\\omega & =2 \pi n \\& =2 \times \frac{22}{7} \times \frac{7}{11}=4 \text { radians } / sec\end{aligned}$
Putting the value of $v$ in equation (1)
$\begin{aligned}v & =4 \sqrt{(1)^2-(0.6)^2} \\& =4 \sqrt{1-0.36}=4 \sqrt{0.64} \\& =4 \times 0.8=3.2 m / s\end{aligned}$
$\begin{aligned}\text { Displacement } y & =A \sin \omega t \\& =1.0 \sin (4 t) \because \omega=4 \text { radian } / sec \\y & =1.0 \sin (4 t) m\end{aligned}$

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