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 of a simple harmonic oscillator is: \[ \omega = \sqrt{\frac{k}{m}}, \] where: \begin{itemize} \item \( k \) is the spring constant, which represents the stiffness of the spring. \item \( m \) is the mass of the oscillator. \end{itemize} Derivation: \begin{itemize} \item The restoring force in SHM is \( F = -kx \), where \( x \) is displacement. \item Using Newton’s second law \( F = ma \), where \( a = \ddot{x} \): \[ m\ddot{x} + kx = 0. \] \item This is a second-order differential equation whose solution gives \( \omega = \sqrt{\frac{k}{m}} \). \end{itemize} ---