Question

The equation of a particle executing simple harmonic motion is \( x = a \sin(\omega t) + b \cos(\omega t) \). The maximum velocity of the particle is:

• A. \( \omega (a + b) \)
• B. \( \omega \left( \sqrt{a} + \sqrt{b} \right) \)
• C. \( \omega \sqrt{a^2 + b^2} \)
• D. \( \frac{\omega (a + b)}{2} \)

Hint:

For a particle executing simple harmonic motion, the maximum velocity is given by \( v_{\text{max}} = \omega \sqrt{a^2 + b^2} \), where \( a \) and \( b \) are the coefficients in the displacement equation.

Answer 1

The Correct Option is C

The displacement equation is \( x(t) = a \sin(\omega t) + b \cos(\omega t) \). The velocity is the derivative of the displacement: \[ v(t) = a \omega \cos(\omega t) - b \omega \sin(\omega t) \] The maximum velocity occurs when the sine and cosine functions are at their maximum values: \[ v_{\text{max}} = \omega \sqrt{a^2 + b^2} \]
Thus, the maximum velocity is \( v_{\text{max}} = \omega \sqrt{a^2 + b^2} \).