University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

297

Pages:

6
Author:

Jabin F.

Oscillation of a Particle in a Tunnel Made Along a Chord of Earth Consider Earth as a uniform solid sphere (MR). In the figure shown, the particle (m) will experience force only due to the core of radius r, not due to the shell R-r. To find the mass of the core (density is equal): (M/4) * π * R³ = (M_core) * (4/3) * π * r³ M_core = M * (r/R)³ Force on the particle due to the core: F = G * m * M_core / r² = G * m / r² * M * (r/R)³ F = G * m * M * r / R³ Equation of Motion (EOM): F * cos θ = m * ẍ (G * m * M * r / R³) * (x / r) = m * ẍ ẍ = - (G * M / R³) * x This is a simple harmonic motion (SHM) for both small and large oscillations. ω = √(G * M / R³) T = 2π * √(R³ / G * M) Acceleration due to gravity of Earth's surface: ẍ = a = g g = G * M / R² GM = g * R² T = 2π * √(R³ / (9 * R²)) = 2π * √(R / 9) P-53: Imagine a narrow tunnel excavated through the Earth as shown in the diagram. A person falls into the tunnel at one end. Assuming the tunnel is frictionless, the person will: (a) fall straight through, escaping Earth's gravity (b) describe simple harmonic motion with period 2π√(d/R) * √(R³/GM) (c) describe simple harmonic motion with period 2π√((R-d)³/GM) (d) describe simple harmonic motion with period 2π√(R³/GM) Solution: As described in previous articles, such type of motion is SHM with period T = 2π√(R³/GM), which is independent of position. P-62 and Q-64: A perfectly straight tunnel is dug between two points on the surface of the Earth. A particle is allowed to slide without friction in this tunnel under the action of Earth's gravity. The particle will execute SHM with period: (a) 3π√(3/ρG) (b) 3√(3/ρG) (c) 3π√(3/5G) (d) 3√(3/5G) Solution: The time period of SHM in this case is T = 2π√(R³/GM). Substituting ρ = M / ((4/3)πR³) into the equation gives T = 3π√(3/ρG). P-5: Consider the Earth as a uniform sphere of total mass M and radius R. A small object slides along a tunnel connecting two points on the surface of the Earth and is acted upon only by the gravitational force due to Earth. Using the value of acceleration due to gravity as 9.81 m/s² and the radius of Earth as 6.37 x 10⁶ m, the time to slide between the two points is: Solution: The time period of oscillation is T = 9π√(R/g). So, time from extreme to extreme is T/2. Calculate the time using the given values. Q-32: In the figure shown, the particle is released at the end of the tunnel. What is its speed at the center of the tunnel? Solution: T = 9π√(R/g) => ω = 2π/T = √(g/R) The speed at the center of the tunnel is maximum, so v = ωR = √(gR) = √(GM/R)