Question

A simple pendulum of length $20\text{cm}$ and mass $5.0\text{g}$ is suspended in a race car travelling with constant speed $70\text{m/s}$ around a circle of radius $50\text{m}$. If the pendulum undergoes small oscillations about its equilibrium position, what is the frequency of oscillation?

• A. $32\text{Hz}$
• B. $25\text{Hz}$
• C. $21\text{Hz}$
• D. $3.5\text{Hz}$

Answer 1

The correct option is D $3.5\text{Hz}$

Since the centripetal acceleration is horizontal and Earth's gravitational $\overrightarrow{g}$ is downward, we can define the magnitude of an "effective" gravitational acceleration using the Pythagoras theorem:
$g_{eff}=\sqrt{g^2+{\left(\frac{v^2}{R}\right)}^2}$.
Then, since frequency is the reciprocal of the period,
$f=\frac{1}{2\pi }\sqrt{\frac{g_{eff}}{L}}=\frac{1}{2\pi }\sqrt{\frac{\sqrt{g^2+\frac{v^4}{R^2}}}{L}}$
With $v=70\text{m/s},R=50\text{m},$ and $L=0.20\text{m},$ we have $f\approx 3.5{\text{s}}^{-1}=3.5\text{Hz}$