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: Approach
Find the amplitude through the maximum of $x$.

Step 2: Maximum
$x=a\sin\omega t+b\cos\omega t$ has a maximum value equal to the length of the vector $(a,b)$, because it is a dot product of $(a,b)$ with the unit vector $(\sin\omega t,\cos\omega t)$. So $x_{max}=\sqrt{a^2+b^2}$.

Step 3: Acceleration check
$\dfrac{d^2x}{dt^2}=-\omega^2x$, which is the SHM condition.

Step 4: Answer
Simple harmonic with amplitude $\sqrt{a^2+b^2}$, option (B).

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