Oscillations - Result Question 3

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

$y=A_0+A \sin \omega t+B \cos \omega t$.

[2019]

Then the amplitude of its oscillation is given by:

(a) $A_0+\sqrt{A^{2}+B^{2}}$

(b) $\sqrt{A^{2}+B^{2}}$

(c) $\sqrt{A_0^{2}+(A+B)^{2}}$

(d) $A+B$

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Answer:

Correct Answer: 4. (b)

Solution:

  1. (b) $B$

Given equation

$y=A_0+A \sin \omega t+B \sin \omega t$

Now assume $(y-A_0)=\gamma$

$y-A_0=A \sin \omega t+B \sin \omega t$

$\gamma=A \sin \omega t+B \cos \omega t$

$=\sqrt{A^{2}+B^{2}} \sin (\omega t+\phi)$

which is S.H.M.

where $\cos$

and $\sin \phi=\frac{B}{\sqrt{A^{2}+B^{2}}}$

so, resultant amplitude

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