Question

The displacement of a particle varies with time according to the relation
\(x = asinωt+bcosωt\)

• A. The motion is simple harmonic motion with an amplitude \(\sqrt[3]{a^2+b^2}\)
• B. The motion is simple harmonic motion with an amplitude \(\sqrt{a^2+b^2}\)
• C. The motion is periodic but not simple harmonic motion
• D. The motion is simple harmonic with an amplitude \((a^2+b^2)\)

Hint:

Combine the sine and cosine into a single sine with an amplitude and phase.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A sum of a sine and a cosine of the same angular frequency is a single sinusoid with the same frequency.

Step 2: Rewrite
Let \(a=A\cos\phi\) and \(b=A\sin\phi\). Then
\[ x=A\cos\phi\sin\omega t+A\sin\phi\cos\omega t=A\sin(\omega t+\phi) \]

Step 3: Find A
\[ a^2+b^2=A^2\Rightarrow A=\sqrt{a^2+b^2} \]

Step 4: Nature of motion
\(x=A\sin(\omega t+\phi)\) is the standard form of simple harmonic motion with amplitude \(A\).

Step 5: Check the options
The cube root and the plain sum \(a^2+b^2\) are not amplitudes. The motion is simple harmonic, not merely periodic. So the answer is (B).

Final Answer:
The expression equals a single sine with amplitude root of a squared plus b squared, so it is SHM, option (B). \[ \boxed{\text{SHM, amplitude }\sqrt{a^2+b^2}} \]