C1-2. Modeling Physical Oscillators

Example 1

Problem: A simple pendulum of length $L$ and a vertical mass-spring system (mass $m$, spring constant $k$) both oscillate with the exact same period $T_0$ on Earth's surface.

Both systems are transferred inside a rocket that launches vertically upward with a constant acceleration of $a = 3g$. While inside the accelerating rocket:

1. The string of the pendulum is shortened to $\dfrac{1}{4}L$.
2. The spring of the mass-spring system is cut exactly in half, and the original mass $m$ is attached to one of the halves.

Derive expressions for the new periods of the pendulum ($T_p$) and the mass-spring system ($T_s$) in terms of $T_0$, and determine which system now oscillates with a higher frequency.

Solution:

1. Analyzing the Pendulum ($T_p$):

• On Earth, the initial period is: $T_0 = 2\pi \sqrt{\dfrac{L}{g}}$
• Inside an upwardly accelerating rocket, an inertial "effective gravity" ($g_{\text{eff}}$) acts downward on the pendulum bob: $$g_{\text{eff}} = g + a = g + 3g = 4g$$
• Substituting the new length $L' = \dfrac{1}{4}L$ and $g_{\text{eff}} = 4g$ into the pendulum period formula: $$T_p = 2\pi \sqrt{\dfrac{\frac{1}{4}L}{4g}} = 2\pi \sqrt{\dfrac{L}{16g}} = \dfrac{1}{4} \left(2\pi \sqrt{\dfrac{L}{g}}\right) = \mathbf{\dfrac{1}{4}T_0}$$

2. Analyzing the Mass-Spring System ($T_s$):

• On Earth, the initial period is: $T_0 = 2\pi \sqrt{\dfrac{m}{k}}$
• The period of a mass-spring system is dictated purely by inertia ($m$) and the restoring stiffness ($k$). Changes in gravity or acceleration shift the static equilibrium position; they have absolutely zero effect on the execution period.
• Spring stiffness is inversely proportional to its length. When a spring is cut precisely in half, each individual segment becomes twice as stiff because it takes twice as much force to stretch the shortened coils by the same absolute displacement. Therefore, the new spring constant is $k' = 2k$.
• Substituting $k' = 2k$ into the mass-spring period formula: $$T_s = 2\pi \sqrt{\dfrac{m}{2k}} = \dfrac{1}{\sqrt{2}} \left(2\pi \sqrt{\dfrac{m}{k}}\right) = \mathbf{\dfrac{1}{\sqrt{2}}T_0}$$

Conclusion & Frequency Comparison:

• Comparing the periods: $T_p = 0.25T_0$ and $T_s \approx 0.707T_0$.
• Since the pendulum has a significantly shorter period ($T_p < T_s$), it completes its cycles faster. Because frequency is the reciprocal of the period ($f = \dfrac{1}{T}$), the pendulum oscillates with a higher frequency ($f_p = 4f_0$ vs $f_s = \sqrt{2}f_0$).