Q.1 (viii) A particle is subjected to two parallel S.H.M.s such that x = 2sinω t and y = 2sin(ω t + π/3) . The amplitude of the resultant S.H.M. will be _____. A. 0 B. 2√(3) C. 4 D. 12

Answer: B. 2√(3) Answer: (b) 2√(3) For two parallel S.H.M.s of the same period, A_R = √(A₁² + A₂² + 2A₁A₂cosδ) . Here A₁ = A₂ = 2 , δ = π/3 : A_R = √(4 + 4 + 2(2)(2)cos 60^°) = √(8 + 4) = √(12) = 2√(3)

A particle is subjected to two parallel S.H.M.s such that \(x = 2\sin\omega t\) and \(y = 2\sin\left(\omega t + \dfrac{\pi}{3}\right)\). The amplitude of the resultant S.H.M. will be _____.

Q.1 (viii) A particle is subjected to two parallel S.H.M.s such that \(x = 2\sin\omega t\) and \(y = 2\sin\left(\omega t + \dfrac{\pi}{3}\right)\). The amplitude of the resultant S.H.M. will be _____.

A. 0

B. \(2\sqrt{3}\)

C. 4

D. 12

Doubtly Asked 27 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: B. \(2\sqrt{3}\)

Answer: (b) \(2\sqrt{3}\)

For two parallel S.H.M.s of the same period, \(A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\delta}\). Here \(A_1 = A_2 = 2\), \(\delta = \pi/3\):

\[A_R = \sqrt{4 + 4 + 2(2)(2)\cos 60^\circ} = \sqrt{8 + 4} = \sqrt{12} = 2\sqrt{3}\]

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

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A particle is subjected to two parallel S.H.M.s such that \(x = 2\sin\omega t\) and \(y = 2\sin\left(\omega t + \dfrac{\pi}{3}\right)\). The amplitude of the resultant S.H.M. will be _____.

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