Solution: Train Oscillating on a Hill
Let \( x \) be the length of the train on the hill at any instant. The retarding force due to gravity acting on this portion is:
$$ F = -(\text{mass of } x) g \sin \theta = -\left( \frac{M}{l} x \right) g \sin \theta $$The equation of motion for the whole train (mass \( M \)) is:
$$ Ma = -\frac{M}{l} g \sin \theta \cdot x \implies a = -\left( \frac{g \sin \theta}{l} \right) x $$This is the equation of SHM with angular frequency \( \omega = \sqrt{\frac{g \sin \theta}{l}} \).
The train enters the hill (starts at \( x=0 \) with max velocity) and stops momentarily when a fraction \( \eta \) has risen (at \( x = \eta l \)). This motion corresponds to moving from the equilibrium position to the extreme position in SHM. The time taken is exactly one quarter of the time period (\( T_{shm}/4 \)). The time to slide back down to the start is also \( T_{shm}/4 \).
Total time spent on the hill:
$$ t = \frac{T_{shm}}{2} = \frac{\pi}{\omega} = \pi \sqrt{\frac{l}{g \sin \theta}} $$Given: \( l = 900 \) m, \( \sin(9.2^\circ) = 0.16 \), \( \sqrt{g} = \pi \) (so \( g \approx \pi^2 \)).
$$ t = \pi \sqrt{\frac{900}{\pi^2 \times 0.16}} = \pi \frac{1}{\pi} \sqrt{\frac{900}{0.16}} $$ $$ t = \sqrt{\frac{900}{0.16}} = \frac{30}{0.4} = 75 \text{ s} $$