Q106 / 180 · NEET 2019 (All India) · Physics

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

A
$A_{0} + \sqrt{A^{2} + B^{2}}$
Correct answer
B
$\sqrt{A^{2} + B^{2}}$
C
$\sqrt{A_{0}^{2} + {(A + B)}^{2}}$
D
$A + B$

Solution

y = A_{0} + A \sin (ω t) + B \cos (ω t)

= A_{0} + \sqrt{A^{2} + B^{2}} \cdot [\frac{A}{\sqrt{A^{2} + B^{2}}} \sin (ω t) + \frac{B}{\sqrt{A^{2} + B^{2}}} \cos (ω t)]

= A_{0} + [\sqrt{A^{2} + B^{2}}] \sin (ω t + φ)

Amplitude

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

Hence, the correct answer is option: 'B'.

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