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STD 11 - 13. oscillations — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 13. oscillations4 Marks
MCQ
A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then, its time period of vibration will be
  • ✓
    $\frac{{2\pi \beta }}{\alpha }$
  • B
    $\frac{{{\beta ^2}}}{{{\alpha ^2}}}$
  • C
    $\frac{\alpha }{\beta }$
  • D
    $\frac{{{\beta ^2}}}{\alpha }$
✓

Answer

Correct option: A.
$\frac{{2\pi \beta }}{\alpha }$
a
If $A$ and $\omega$ be amplitude and angular frequency of vibration, then

${\alpha=\omega^{2} A} ......... (i)$

${\text { and } \beta=\omega A} ......... (ii)$

Dividing eqn. $(i)$ by eqn. $(ii),$ we get

$\frac{\alpha}{\beta}=\frac{\omega^{2} A}{\omega A}=\omega$

$\therefore$ Time period of vibration is

$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{(\alpha / \beta)}=\frac{2 \pi \beta}{\alpha}$

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