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PART - 2 CH - 13 Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 13 Oscillations2 Marks
Question
A particle performing simple harmonic motion makes 1200 oscillations per minute and its velocity while passing through the mean position is $3.14 m / s$ lives. Also obtain the displacement equation of the particle if the displacement at moment $t=0$.
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Answer

We know that the velocity of the particle is maximum in the mean position. If the particle's amplitude is $A$ and angular velocity is $\omega$ then maximum velocity.
$\begin{aligned}v_{\max } & =\omega A \\\text { Amplitude } A & =\frac{V_{\max }}{\omega}=\frac{V_{\max }}{2 \pi n}\end{aligned}$
$\because \omega=2 \pi n$, Here frequency is $n$,
As per question
$\begin{aligned}v_{\max } & =3.1 m / s \\n & =1200 \text { per minute }=\frac{1200}{60} \text { per second } \\n & =20 \text { per second } \\\text { Amplitude } A & =\frac{3.1}{2 \times 3.14 \times 20}=\frac{1}{40}=0.025 m\end{aligned}$
Displacement of the particle at $t=0$ that is the particle is in its equilibrium state, then the equation of simple harmonic motion is,
$y=A \sin \omega t=A \sin 2 \pi n t$
Keeping the value of A and $n$
$\begin{array}{l}y=0.025 \sin 2 \pi \times 20 t \\y=0.025 \sin 40 \pi t\end{array}$

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