Question

Simple pendulums, each of length l and mass m, are coupled by a spring with spring constant k. If either of the pendulums is clamped, the angular frequency of the other is given by:

a) √g/l​​
b) √k/m​​
c) √k/2m​​
d) √k/m​+g/l​​

Answer 1

The correct answer for the angular frequency of a simple pendulum coupled by a spring when one of the pendulums is clamped is b) √(k/m), assuming no damping is present.

Step-by-step explanation:

The question pertains to the angular frequency of a simple pendulum coupled by a spring when one of the pendulums is clamped. For small angular displacements, a simple pendulum can be considered a simple harmonic oscillator (SHM), where the restoring force is proportional to the displacement. The expression for angular frequency for a simple pendulum is given by √(g/l), where g is the acceleration due to gravity and l is the length of the pendulum.

However, when a spring with spring constant k is added, and one of the pendulums is clamped, causing the system to behave like a mass-spring oscillator, the angular frequency changes. The angular frequency ω for a mass-spring system in simple harmonic motion is given by ω = √(k/m), assuming no damping. Here, m is the mass attached to the spring. If damping is negligible, the correct answer from the options given would be b) √(k/m).