The displacement of a particle executing simple harmonic motion is given by, y = A _ 0 + Asin + Bcos . Then, the amplitude of its oscillation is given by

Correct answer: B. A ^ 2 + B ^ 2 y = A _ 0 + Asin( ) + Bcos( ) = A _ 0 + A ^ 2 + B ^ 2 · [ A A ^ 2 + B ^ 2 ( ) + B A ^ 2 + B ^ 2 ( )] = A _ 0 + [ A ^ 2 + B ^ 2 ] ( + φ) Amplitude = A ^ 2 + B ^ 2

NEET 2019

Exam: NEET
Year: 2019
Subject: Physics
Chapter: Oscillations

NEET 2019 · Physics · Oscillations

The displacement of a particle executing simple harmonic motion is...

The displacement of a particle executing simple harmonic motion is given by,

y = {A}_{0} + Asin\omegat + Bcos\omegat .

Then, the amplitude of its oscillation is given by

Options

A ${A}_{0} + \sqrt{{A}^{2} + {B}^{2}}$
B $\sqrt{{A}^{2} + {B}^{2}}$
C\sqrt{{A}_{0}^{2} + { $\left(A + B\right)}^{2}}$
D $A + B$

Correct answer

B. $\sqrt{{A}^{2} + {B}^{2}}$

Step-by-step solution

y = {A}_{0} + Asin\left(\omegat\right) + Bcos\left(\omegat\right)

= {A}_{0} + \sqrt{{A}^{2} + {B}^{2}} \cdot \left[\dfrac{A}{\sqrt{{A}^{2} + {B}^{2}}}\sin \left(\omegat\right) + \dfrac{B}{\sqrt{{A}^{2} + {B}^{2}}}\cos \left(\omegat\right)\right]

= {A}_{0} + \left[\sqrt{{A}^{2} + {B}^{2}}\right]\sin \left(\omegat + \phi\right)

Amplitude

= \sqrt{{A}^{2} + {B}^{2}}

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