Respuesta :
Answer:
To find the angular frequency \( \omega \) for small amplitude oscillations of the system, we can use the formula for the angular frequency of a physical pendulum combined with springs:
\[ \omega = \sqrt{\frac{g}{L_{\text{eff}}}} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)) and \( L_{\text{eff}} \) is the effective length of the pendulum. The effective length is influenced by the springs and can be calculated by considering the additional restoring force provided by the springs when the pendulum is displaced.
The force from one spring when the pendulum is displaced by a small angle \( \theta \) is \( -kx \), where \( x \) is the displacement of the spring. Since there are two springs, the total restoring force from the springs is \( -2kx \). The displacement \( x \) can be related to the length \( L \) and the angle \( \theta \) by \( x = L\theta \).
The torque \( \tau \) due to the springs is \( \tau = -2kxL = -2kL^2\theta \), and this torque contributes to the effective restoring torque that determines the angular frequency.
The moment of inertia \( I \) of the mass with respect to the axis of rotation is \( mL^2 \). The equation of motion for the pendulum is then:
\[ I\frac{d^2\theta}{dt^2} = -mgL\theta - 2kL^2\theta \]
For small oscillations, this simplifies to:
\[ \frac{d^2\theta}{dt^2} + \left(\frac{mgL}{mL^2} + \frac{2kL^2}{mL^2}\right)\theta = 0 \]
\[ \frac{d^2\theta}{dt^2} + \left(\frac{g}{L} + \frac{2kL}{m}\right)\theta = 0 \]
The term in parentheses is the square of the angular frequency \( \omega^2 \):
\[ \omega^2 = \frac{g}{L} + \frac{2kL}{m} \]
\[ \omega = \sqrt{\frac{g}{L} + \frac{2kL}{
