First slide

Simple hormonic motion

Question

The displacement of a particle executing simple harmonic motion is given by $y = A_{0} + Asin ωt + B \cos ωt$ . Then the amplitude of its oscillation is given by

Moderate

A

A+B
B

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

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

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

Solution

$y = A_{0} + (A \sin ωt + B \cos ωt)$
= $A_{0} + A \sin ω t + B \sin (ω t + \frac{π}{2})$
Hence 2 SHMs are superimposed with phase difference of $\frac{π}{2}$
Amplitude a = $\sqrt{A^{2} + B^{2} + 2 ABcos ∆ \emptyset}$
As, $∆ \emptyset = \frac{π}{2} Hence a = \sqrt{A^{2} + B^{2}}$
$y = A_{0} + \sqrt{A^{2} + B^{2}} \sin (ωt 6 ϕ)$
$A_{0} is mean position, and \sqrt{A^{2} + B^{2}} is amplitude$

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