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Simple Harmonic Motion — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsSimple Harmonic Motion5 Marks
Question
Consider a particle moving in simple harmonic motion according to the equation $\text{x}=2.0\cos(50\pi\text{t}+\tan^{-1}0.75)$ where x is in centimetre and t in second. The motion is started at $t = 0.$
  1. When does the particle come to rest for the first time?
  2. When does the acceleration have its maximum magnitude for the first time?
  3. When does the particle come to rest for the second time?
✓

Answer

  1. $\text{x}=2.0\cos(50\pi\text{t}+\tan^{-1}0.75)$
$=2.0\cos(50\pi\text{t}+0.643)$
$\text{v}=\frac{\text{dx}}{\text{dt}}=-100\sin(50\pi\text{t}+0.643)$
$\Rightarrow\sin(50\pi\text{t}+0.643)=0$
As the particle comes to rest for the $1^{st}$ time
$\Rightarrow50\pi\text{t}+0.643=\pi$
$\Rightarrow\text{t}=1.6\times10^{-2}\sec.$
  1. Acceleration $\text{a}=\frac{\text{dv}}{\text{dt}}=-100\pi\times50\pi\cos(50\pi\text{t}+0.643)$
For maximum acceleration $\cos(50\pi\text{t}+0.643)=-1\cos\pi$ (max) (so a is max)
$\Rightarrow\text{t}=1.6\times10^{-2}\sec.$
  1. When the particle comes to rest for second time,
$50\pi\text{t}+0.643=2\pi$
$\Rightarrow\text{t}=3.6\times10^{-2}\text{s}.$

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