The displacement of a particle executing simple harmonic motion is given by y - A 0- A sin- t - B cos- Then- the amplitude of its oscillation is given byA- A 2- B 2B- A 0- A 2- B 2C- A - BD- A 02- A - B 2

The correct option is A A2+B2Given, y=A0+Ay = A_0 + A sin ωtωt + B cos ωtωt Equation of nbsp;SHM y' = y A_0 = A sin ω nbsp;t + B cos ω t Resultant ampli ...

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Standard XII

Physics

SHM expression

The displacem...

Question

The displacement of a particle executing simple harmonic motion is given by
$y = A_{0} + A sin ω t + B cos ω t$ . Then, the amplitude of its oscillation is given by

A_{0} + \sqrt{A^{2} + B^{2}}

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

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\sqrt{{A_{0}}_{0}^{2} + {(A + B)}^{2}}

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A + B

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Solution

The correct option is B $\sqrt{A^{2} + B^{2}}$
Given,
$y = A_{0} + A$ sin $ω t$ + $B$ cos $ω t$
Equation of SHM
$y^{'} = y - A_{0} = A$ sin $ω t$ + $B$ cos $ω t$
Resultant amplitude,
$R = \sqrt{A^{2} + B^{2} + 2 A B cos 90 °}$
$= \sqrt{A^{2} + B^{2}}$

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The displacement of a particle executing simple harmonic motion is given by y = A0 + A sinωt + B cosωt. Then, the amplitude of its oscillation is given by

The correct option is B √A2+B2Given, y=A0+A sin ωt + B cos ωt Equation of SHM y′=y−A0=A sin ωt + B cos ωt Resultant amplitude, R=√A2+B2+2ABcos90° =√A2+B2

The displacement of a particle executing simple harmonic motion is given by
$y = A_{0} + A sin ω t + B cos ω t$ . Then, the amplitude of its oscillation is given by

The correct option is B $\sqrt{A^{2} + B^{2}}$
Given,
$y = A_{0} + A$ sin $ω t$ + $B$ cos $ω t$
Equation of SHM
$y^{'} = y - A_{0} = A$ sin $ω t$ + $B$ cos $ω t$
Resultant amplitude,
$R = \sqrt{A^{2} + B^{2} + 2 A B cos 90 °}$
$= \sqrt{A^{2} + B^{2}}$