WBJEE · Physics · Oscillations
The displacement of a particle in S.H.M. varies according to the relation \(x=4(\cos \pi t+\sin \pi \mathrm{t})\). The amplitude of the particle is
- A -4
- B 4
- C \(4 \sqrt{2}\)
- D 8
Answer & Solution
Correct Answer
(C) \(4 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Hints: \(\mathrm{R} \sin \delta=4\) \(\mathrm{R} \cos \delta=4\) \(R=4 \sqrt{2}\)
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