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NEET
Exam
NEET
Subject
Physics
Chapter
Oscillations

NEET · Physics · Oscillations

For a particle executing simple harmonic motion, the...

For a particle executing simple harmonic motion, the displacement-time (x−t)(x-t) graph is as shown in the figure. The acceleration of the particle at t=43 st=\frac{4}{3} \mathrm{~s} is For a particle executing simple harmonic motion, the displacement-time (x-t) graph is as shown in the figure. The

Options

  1. A−332π2 cm s−2-\frac{\sqrt{3}}{32} \pi^2 \mathrm{~cm} \mathrm{~s}^{-2}
  2. B323π2 cm s−2\frac{32}{\sqrt{3}} \pi^2 \mathrm{~cm} \mathrm{~s}^{-2}
  3. C+332πcms−2+\frac{\sqrt{3}}{32} \pi \mathrm{cm} \mathrm{s}^{-2}
  4. D+323πcms−2+\frac{32}{\sqrt{3}} \pi \mathrm{cm} \mathrm{s}^{-2}

Correct answer

A. −332π2 cm s−2-\frac{\sqrt{3}}{32} \pi^2 \mathrm{~cm} \mathrm{~s}^{-2}

Step-by-step solution

The displacement-time graph shown in the figure is a sine wave, so the equation of displacement, x=1sin⁡ωtx=1 \sin \omega t For a particle executing simple harmonic motion, the displacement-time (x-t) graph is as shown in the figure. The
Here, T=8s\quad T=8 \mathrm{ s} (for a complete cycle) Hence, ω=2πT=2π8=π4rad/sandaccelerationinSHM,\omega=\frac{2 \pi}{T}=\frac{2 \pi}{8}=\frac{\pi}{4} \mathrm{rad} / \mathrm{s} and acceleration in SHM,α=−ω2asin⁡ωtα=−(π4)2×1×sin⁡(π4×43)⇒α=−π216×32=−π2332cms−2\begin{aligned} \alpha& =-\omega^2 a \sin \omega t \\ \alpha& =-\left(\frac{\pi}{4}\right)^2 \times1 \times \sin \left(\frac{\pi}{4} \times \frac{4}{3}\right)\\ \Rightarrow \quad \alpha& =-\frac{\pi^2}{16} \times \frac{\sqrt{3}}{2}=-\frac{\pi^2 \sqrt{3}}{32} \mathrm{ cm} \mathrm{ s}^{-2} \end{aligned}$ Hence, the correct option is (a).
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