Question 221: Oscillations

Question

A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m . When the particle passes through the mean position, its kinetic energy is $16 	imes 10^{-3} \mathrm{~J}$. The equation of motion of this particle is (Initial phase of oscillation $=0^{\circ}$ )

Accepted Answer

At the mean position kinetic energy is maximum which is also equal to the total energy of the S.H.M.
$$ \begin{aligned} & 	herefore \quad \frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2=16 	imes 10^{-3} \ & \Rightarrow \quad \frac{1}{2} 	imes 0.2 	imes \omega^2 	imes(0.2)^2=16 	imes 10^{-3} \ & \Rightarrow \quad \omega^2=\frac{16 	imes 10^{-3}}{0.004}=4 \ & \Rightarrow \quad \omega=2 \mathrm{rad} / \mathrm{s} \ &\,\,\,\,\,\,\,\,  \quad y=A \sin \omega t \ & \Rightarrow \quad y=0.2 \sin (2 t) \end{aligned} $$

Answer

$\mathrm{Y}=0.2 \sin (4 \mathrm{t})$

Answer

$\mathrm{Y}=0.2 \sin \left(\frac{\mathrm{t}}{4}\right)$

Answer

$\mathrm{Y}=0.2 \sin \left(\frac{\mathrm{t}}{2}\right)$

Answer

$\mathrm{Y}=0.2 \sin (2 \mathrm{t})$

Oscillations

A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m . When the particle passes through the mean position, its kinetic energy is 16 \times 10^{-3} \mathrm{~J}. The equation of motion of this particle is (Initial phase of oscillation =0^{\circ} )

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