Verified Solution • 2025–26 Session moderate

QNEET-2019-Ph-815: NEET 2019 Physics - Class 11 NEET-2019-Ph-815 Solution: The displacement of a particle executing simple harmonic motion …

The displacement of a particle executing simple harmonic motion is given by y=A0+Asin⁡ωt+Bcos⁡ωty = A_0 + A\sin\omega t + B\cos\omega t. Then the amplitude of its oscillation is given by: A2+B2\sqrt{A^2+B^2} A+BA+B A+A2+B2\sqrt{A+A^2+B^2} A02+(A+B)2\sqrt{A_0^2+(A+B)^2}

Multiple Choice Options

  • A A2+B2\sqrt {{A^2} + {B^2}} A 2 + B 2 _
  • B A + B
  • C A + A2+B2\sqrt {{A^2} + {B^2}} A 2 + B 2 _
  • D A02+(A+B)2\sqrt {A_0^2 + {{\left( {A + B} \right)}^2}} A 0 2 _ + ( A + B ) 2 _

Step-by-Step Verified Solution

From the given displacement y = A 0 + A sin _ t + B cos _ t. Let assume, y - A 0 = _ _ = A sin _ t + B cos _ t = A 2 + B 2 sin _ ( _ t + _ ) which is S.H.M _ Resultant amplitude of the particle, = A 2 + B 2

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