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 find the amplitude of the particle executing simple harmonic motion, we consider the given equation for displacement: \(y = A_0 + A \sin \omega t + B \cos \omega t\).

The general form for a simple harmonic motion (SHM) displacement is:

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

Here, \(C\) is the amplitude of the motion.

We need to express the given equation in this form. We can use the approach of expanding the general equation using trigonometric identities:

• Rewrite \(A \sin \omega t + B \cos \omega t\) as a single trigonometric term:
• We recognize that it can be rewritten as: \(\sqrt{A^2 + B^2} \sin(\omega t + \phi)\)
• Where \(\tan \phi = \frac{B}{A}\), and the amplitude \(C\) of this trigonometric term is \(\sqrt{A^2 + B^2}\).

Notice that the constant \(A_0\) does not affect the amplitude of these oscillations. Amplitude is determined by the maximum value of the oscillating part, which is \(\sqrt{A^2 + B^2}\).

Therefore, the amplitude of the oscillation is indeed given by:

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

Thus, the correct answer is option 2: \(\sqrt {A^2+B^2}\).