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Oscillations question

2019 · Q154
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Oscillations question

2019 · Q154

NEETPhysicsOscillationsMCQ+4 / −1
The displacement of a particle executing simple harmonic motion is given by

y = A0 + A sinω\omegaωt + B cosω\omegaωt.

Then the amplitude of its oscillation is given by :
  1. A
    A2+B2\sqrt {{A^2} + {B^2}}A2+B2​
  2. B
    A + B
  3. C
    A + A2+B2\sqrt {{A^2} + {B^2}}A2+B2​
  4. D
    A02+(A+B)2\sqrt {A_0^2 + {{\left( {A + B} \right)}^2}}A02​+(A+B)2​
View written solutionFree

Correct answer: A

From the given displacement

y = A0 + A sinω\omega ωt + B cosω\omega ωt.

Let assume, y - A0 = γ\gamma γ

γ\gamma γ = A sinω\omega ωt + B cosω\omega ωt

= A2+B2sin⁡(ωt+ϕ)\sqrt {{A^2} + {B^2}} \sin \left( {\omega t + \phi } \right)A2+B2​sin(ωt+ϕ)

which is S.H.M

∴\therefore∴ Resultant amplitude of the particle,

= A2+B2\sqrt {{A^2} + {B^2}} A2+B2​

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