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Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsOscillations5 Marks
Question
A point particle of mass $0.1kg$ is executing S.H.M. of amplitude of $0.1m$. When the particle passes through the mean position, its kinetic energy is $8 \times 10^{-3}$ joule. Obtain the equation of motion of this particle if the initial phase of oscillation is $45^\circ$
✓

Answer

The displacement of a particle in S.H.M is given by$\text{y}=\text{a}\sin(\omega\text{t}+\phi)$
Velocity $=\frac{\text{dy}}{\text{dt}}=\omega\text{a}\cos(\omega\text{t}+\phi)$ The velocity is maximum when the particle passes through the mean position i.e.$=\Big(\frac{\text{dy}}{\text{dt}}\Big)_\text{max}=\omega\text{a}$
The kinetic energy at this instant is given by$=\frac{1}{2}\text{m}\Big(\frac{\text{dy}}{\text{dt}}\Big)^2$
$=\frac{1}{2}\text{m}\times\omega^2\text{a}^2$
$=8\times10^{-3}\text{ joule}$
$\frac{1}{2}\times(0.1)\omega^2\times(0.1)^2$
$=8\times10^{-3}\text{ joule}$
Solving we get $\omega=\pm4$ Substituting the values of a $\omega$ and $\phi$ in the equation of SHM., we get$\text{y}=0.1\sin(\pm4\text{t}+\frac{\pi}{4})\text{ metre}$

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