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JEE Main 2025 (2 Apr, Shift 1) · Damped, Forced Oscillations & Superposition

Damped, Forced Oscillations & Superposition question from JEE Main 2025 (2 Apr, Shift 1)

Q2JEE Main 2025 · 2 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition

A particle is subjected to two simple harmonic motions as: x1=7sin⁡5t cmx_1 = \sqrt{7}\sin 5t\text{ cm} and x2=27sin⁡(5t+π3) cmx_2 = 2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)\text{ cm} where xx is displacement and tt is time in seconds. The maximum acceleration of the particle is x×10−2 m s−2x \times 10^{-2}\text{ m s}^{-2}. The value of xx is:

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Answer: (d)

  1. Resultant amplitude: A2=A12+A22+2A1A2cos⁡π3A^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\frac{\pi}{3}.
  2. A2=7+28+2(7)(27)(12)=49⇒A=7 cmA^2 = 7 + 28 + 2(\sqrt{7})(2\sqrt{7})\left(\frac{1}{2}\right) = 49 \Rightarrow A = 7\ \text{cm}.
  3. amax=ω2A=25×7=175 cm s−2=175×10−2 m s−2a_{max} = \omega^2 A = 25 \times 7 = 175\ \text{cm s}^{-2} = 175 \times 10^{-2}\ \text{m s}^{-2}.
  4. So the answer is (d) 175.
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