Question

What is the angular frequency \( \omega \) of a simple harmonic oscillator with mass \( m \) and spring constant \( k \)?

• A. \( \sqrt{\frac{k}{m}} \)
• B. \( \sqrt{\frac{m}{k}} \)
• C. \( \frac{k}{m} \)
• D. \( \frac{m}{k} \)

Hint:

Remember, the time period \( T \) is related to angular frequency as \( T = \frac{2\pi}{\omega} \).

Answer 1

The Correct Option is A

The angular frequency \( \omega \) of a simple harmonic oscillator is determined by the equation: \[ \omega = \sqrt{\frac{k}{m}}, \] where \( k \) signifies the spring constant, indicating the spring's rigidity, and \( m \) represents the oscillator's mass. Derivation: The process begins with the restoring force in simple harmonic motion, given by \( F = -kx \), where \( x \) is the displacement. Applying Newton's second law, \( F = ma \), with \( a = \ddot{x} \), yields: \[ m\ddot{x} + kx = 0. \] This second-order differential equation's solution directly provides the angular frequency as \( \omega = \sqrt{\frac{k}{m}} \).