Question

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

• A. $\sqrt {A^2_0 +(A+B)^2}$
• B. $A+B$
• C. $A_0 +\sqrt{A^2+B^2}$
• D. $\sqrt{A^2+B^2}$

Answer 1

The Correct Option is D

The problem involves determining the amplitude of a particle executing simple harmonic motion (SHM) described by the displacement equation:

y = A_{0} + A \sin \omega t + B \cos \omega t

To find the amplitude of oscillation, consider the general form of SHM given by:

y = C \sin (\omega t + \phi)

where C represents the amplitude and \phi is the phase angle.

In the provided displacement equation, the time-dependent part is:

y' = A \sin \omega t + B \cos \omega t

This can be rewritten in the form of a single sine function:

y' = R \sin(\omega t + \phi)

where R = \sqrt{A^2 + B^2} is the amplitude, and \phi is a phase constant.

The given equation is a superposition of sine and cosine functions. To find the resultant amplitude, we apply the relation:

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

This is because if two perpendicular components A \sin \omega t and B \cos \omega t are combined, the amplitude is determined by the Pythagorean theorem.

Therefore, the amplitude of the oscillation described by the displacement equation is:

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

The solution can also be derived geometrically by visualizing the A \sin \omega t and B \cos \omega t components as orthogonal vectors in a plane. The resultant vector, representing the amplitude, has a magnitude given by this expression.

Thus, the correct answer to the question is:

\sqrt{A^2 + B^2}