Question

The time period of the inertial oscillation at a location R is 1.5 times that of a particle moving at a speed of \(0.5~\mathrm{m/s}\) at a location S (87°E, 45°S). Which of the following is the latitude of location R? (Round off to the nearest integer)

• A. 45°N
• B. 67°S
• C. 28°N
• D. 50°S

Hint:

Inertial period \(T \propto 1/\sin\phi\). Longer periods occur at lower latitudes; near the equator, inertial period tends to infinity since \(\sin\phi \to 0\).

Answer 1

The Correct Option is C

Step 1: Formula for inertial period.
The inertial oscillation period is: \[ T = \frac{2\pi}{f}, f = 2\Omega \sin\phi, \] where \(f\) is the Coriolis parameter, \(\Omega = 7.292 \times 10^{-5}\ \mathrm{rad/s}\), and \(\phi\) is latitude.

Step 2: Relating two locations.
\[ \frac{T_R}{T_S} = \frac{f_S}{f_R} = \frac{\sin\phi_S}{\sin\phi_R}. \] Given: \(T_R = 1.5 T_S\). So, \[ 1.5 = \frac{\sin\phi_S}{\sin\phi_R}. \]

Step 3: Substitution for location S.
At location S: \(\phi_S = 45^\circ S\). So, \(\sin\phi_S = \sin 45^\circ = 0.7071\). \[ 1.5 = \frac{0.7071}{\sin\phi_R} \Rightarrow \sin\phi_R = \frac{0.7071}{1.5} = 0.4714. \]

Step 4: Find latitude.
\(\phi_R = \arcsin(0.4714) \approx 28^\circ\).



Step 5: Select hemisphere.
Options given include 28°N, not 28°S. Both are possible mathematically, but usually inertial oscillations are symmetric in both hemispheres. Since option 28°N is listed, that is the correct choice. Final Answer:
\[ \boxed{28^\circ N} \]