Oscillations

JEE Main level, single correct, medium

70 of 194

Question 70 of 194

The displacement of a particle varies according to the relation x=4(cos⁡πt+sin⁡πt)x=4(\cos \pi t+\sin \pi t). The amplitude of the particle is

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Answer: (C) 424 \sqrt{2}

x=4(cos⁡πt+sin⁡πt)=4[sin⁡(π2−πt)]+sin⁡πt]=4[2×sin⁡(πt−π2−πt2)cos⁡(πt−π2+πt2)]\left.x=4(\cos \pi t+\sin \pi t)=4\left[\sin \left(\frac{\pi}{2}-\pi t\right)\right]+\sin \pi t\right]=4\left[2 \times \sin \left(\frac{\pi t-\frac{\pi}{2}-\pi t}{2}\right) \cos \left(\frac{\pi \mathrm{t}-\frac{\pi}{2}+\pi \mathrm{t}}{2}\right)\right]
=8[sin⁡π4⋅cos⁡(−π4+πt)]=8\left[\sin \frac{\pi}{4} \cdot \cos \left(-\frac{\pi}{4}+\pi t\right)\right]
=82⋅cos⁡[πt−π4]=42cos⁡[πt−π4]=\frac{8}{\sqrt{2}} \cdot \cos \left[\pi t-\frac{\pi}{4}\right]=4 \sqrt{2} \cos \left[\pi t-\frac{\pi}{4}\right]
Comparing it with standard equation
X=Acos⁡(wt−Kx)⇒A=42X=A \cos (w t-K x) \quad \Rightarrow A=4 \sqrt{2}

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