Question

The displacement of a particle executing simple harmonic motion is given by \(y=A_0+A sin\omega t+Bcos\omega t.\) Then the amplitude of its oscillation is given by:

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

Answer 1

The Correct Option is B

To determine the amplitude of the given simple harmonic motion, we start by examining the expression for the displacement:

\(y = A_0 + A \sin \omega t + B \cos \omega t\)

We want the amplitude of the oscillation, which is characterized by the part of the expression that is oscillatory, i.e., the terms involving trigonometric functions.

Consider the expression:

\(Y = A \sin \omega t + B \cos \omega t\)

This can be rewritten as a single sinusoidal function:

\(Y = R \sin(\omega t + \phi)\)

To find \(R\), which is the amplitude of the oscillation, we use the trigonometric identity:

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

Therefore, the amplitude of the oscillatory part of the motion is \(\sqrt{A^2 + B^2}\). This is the result of combining the sinusoidal components into a single amplitude.

The term \(A_0\) represents a constant offset and does not affect the amplitude of the oscillation.

The correct answer is: \(\sqrt{A^2 + B^2}\)