A particle is subjected to two simple harmonic motions: x₁ = √7 sin(5t) cm and x₂ = 2√7 sin(5t + π/3) cm. The maximum acceleration is x×10⁻² m s⁻². The value of x is:
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Question

A particle is subjected to two simple harmonic motions: x₁ = √7 sin(5t) cm and x₂ = 2√7 sin(5t + π/3) cm. The maximum acceleration is x×10⁻² m s⁻². The value of x is:

G1 Physics · Accepted Answer

The two SHMs have the same angular frequency ω=5 rad/s. Their amplitudes are A₁=√7 cm, A₂=2√7 cm, and phase difference φ=π/3. Resultant amplitude: A = √(A₁² + A₂² + 2A₁A₂cosφ) = √[7 + 28 + 2(√7)(2√7)(1/2)] = √(7+28+14) = √49 = 7 cm = 0.07 m. Maximum acceleration: a(max) = ω²A = 25×0.07 = 1.75 m s⁻² = 175×10⁻² m s⁻². Hence x = 175. Correct option: B.