AP EAMCET · PHYSICS · Oscillations
If the displacement \(y~(\mathrm{in~} \mathrm{cm})\) of a particle executing simple harmonic motion is given by the equation \(y=5 \sin (3 \pi t)+5 \sqrt{3} \cos (3 \pi t)\), then the amplitude of the particle is
- A 5 cm
- B \(5 \sqrt{3} \mathrm{~cm}\)
- C \(5(1+\sqrt{3}) \mathrm{cm}\)
- D 10 cm
Answer & Solution
Correct Answer
(D) 10 cm
Step-by-step Solution
Detailed explanation
\(A = \sqrt{A_1^2 + A_2^2}\) \(A = \sqrt{(5)^2 + (5\sqrt{3})^2}\) \(A = \sqrt{25 + 75}\) \(A = \sqrt{100}\) \(A = 10 \mathrm{~cm}\)
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