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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 13. oscillations4 Marks
MCQ
A particle is executing $SHM$ with amplitude $A,$ time period $T,$ maximum acceleration $a_o$ and maximum velocity $v_0.$ Its starts from mean position at $t=0$ and at time $t,$ it has the displacement $A/2,$ acceleration $a$ and velocity $v$ then
  • A
    $t=T/12$
  • B
    $a=a_o/2$
  • C
    $v=v_o/2$
  • ✓
    $(A)$ and $(B)$ both
✓

Answer

Correct option: D.
$(A)$ and $(B)$ both
d
Let the displacement at t is $x=A \sin \omega t=A \sin \left(\frac{2 \pi}{T}\right) t$

here, $x=\frac{A}{2}=A \sin \left(\frac{2 \pi}{T}\right) t$

or, $\sin \left(\frac{2 \pi}{T}\right) t=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow t=\frac{T}{12}$

$\frac{d x}{d t}=\frac{2 \pi}{T} A \cos \left(\frac{2 \pi}{T}\right) t$

$\frac{d^{2} x}{d t^{2}}=-\left(\frac{2 \pi}{T}\right)^{2} A \sin \left(\frac{2 \pi}{T}\right) t$

so, $v_{\max }=v_{0}=\left(\frac{2 \pi}{T}\right) A, a_{\max }=a_{0}=\left(\frac{2 \pi}{T}\right)^{2} A$

now at $t=T / 12,$ the velocity, $v=\frac{d x}{d t}=\frac{2 \pi}{T} A \cos \left(\frac{2 \pi}{T}\right) \frac{T}{12}$

$=\frac{2 \pi}{T} A \cos \frac{\pi}{6}=\frac{2 \pi}{T} A \frac{\sqrt{3}}{2}$

$\therefore v=\sqrt{3} v_{0}$

at $t=T / 12,$ the acceleration, $a=\frac{d^{2} x}{d t^{2}}=\left(\frac{2 \pi}{T}\right)^{2} A \sin \left(\frac{2 \pi}{T}\right) \frac{T}{12}$

$=\frac{2 \pi}{T} A \sin \frac{\pi}{6}=\frac{2 \pi}{T} A \frac{1}{2}$

$\therefore a=\frac{a_{0}}{2}$

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