Let $\omega_1, \omega_2$ and $\omega_3$ be the angular speed of the second hand, minute hand and hour hand…

Let $\omega_1, \omega_2$ and $\omega_3$ be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If $x_1, x_2$ and $x_3$ are their respective angular distances in 1 minute then the factor which remains constant $(k)$ is
  1. $\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=k$
  2. $\omega_1 x_1=\omega_2 x_2=\omega_3 x_3=k$
  3. $\omega_1 x_1^2=\omega_2 x_2^2=\omega_3 x_3^2=k$
  4. $\omega_1^2 x_1=\omega_2^2 x_2=\omega_3^2 x_3=k$

Solution

$\omega_1=\frac{2 \pi}{60} ; \quad x_1=\frac{2 \pi}{60} \times 60=2 \pi$ $\omega_2=\frac{2 \pi}{3600} ; \quad x_2=\frac{2 \pi}{3600} \times 60=\frac{2 \pi}{60}$ $\omega_3=\frac{2 \pi}{3600 \times 12} ; \quad x_3=\frac{2 \pi}{3600 \times 12} \times 60=\frac{2 \pi}{720}$ $\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=\frac{1}{60}=k$

Asked in: NEET 2024 (Re-NEET)

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