1. Chapter 7 Models for Wave and Oscillations Variable Gravitational acceleration –Lunar Lander –Escape velocity Wave motion and interaction –Pendulum without damping –Pendulum with damping Mechanical variation

2. Variable Gravitational acceleration Motion with constant gravitational acceleration: motion remains in the immediate vicinity of the earth’s surface –Horizontal motion: Train, MRT, walk & run, ship, ….. –Vertical motion with lower speed: Airplane, jump from an airplane, bullet, ….. Radius of the earth: 6378(Km) & height of airplane: 10 (km) Motion with variable Gravitational acceleration: a projectile in vertical motion doesn’t remains in the immediate vicinity of the earth’s surface –Rocket, Space-shuttle –Missile –Apollo lunar lander –Satellite, ……

3. Variable Gravitational acceleration Problem: The displacement traveled in vertical direction is comparable with the radius of the earth. Newton’s law of gravitation: The gravitational force of attraction between two point masses M and m located at distance r apart is given by –G is a certain empirical constant –The formula is also valid if either or both of the two masses are homogeneous spheres; in this case, the distance r is measured between the center of the spheres.

4. Example: A lunar lander The problem: A lunar lander is free-falling toward the moon’s surface at a speed of v 0 =450 m/s (that is, 1620 km/h). Its retrorockets, when fired in free space, provided a deceleration of T=4 m/s2. At what height above the lunar surface should the retrorockets be activated to ensure a ``soft touchdown’’ (v=0 at impact)? –Mass of the moon: M=7.35E22 (kg) –Radius of the moon: R = 1.74E6 (m) Solution: –let r(t) denote the lander’s distance from the center of the moon at time t –When we combine the (positive) thrust acceleration T and the (negative) lunar acceleration F/m, we get

5. Example: A lunar lander The second order ODE: The problem: Re-write in terms of the lander’s velocity: Substitution using the chain rule formula

6. Example: A lunar lander Integration of both sides with respect to r From the ``soft touchdown’’ condition The solution: Solution: –Find r when v=-450m/s with –T=4m/s2, G=6.6726E-11, M=7.35E22, R=1.74E6

7. Example: A lunar lander The equation for r: The two roots: The result: The lander’s desired initial height above the lunar surface is

8. Escape velocity In 1865, Jules Verne (in his novel from the earth to the moon) raised the question: –What is the initial velocity necessary for a projectile fired from the surface of the earth to reach the moon?? –A similar question: What is the initial velocity necessary for the projectile to escape from the earth altogether?? Condition: If the velocity v remains positive for all t>0, so it continuous forever to move away from the earth!! Variables: –r(t): the projectile’s distance from the earth’s center at time t

9. Escape velocity The equation –Parameters: Mass of the earth: M=5.975E24 (kg) Radius of the earth: R=6.387E6 (m) The constant: G=6.6726E-11 N(m/kg)2 Substitution using the chain rule formula

10. Escape velocity Integration of both sides with respect to r Initial condition The constant C: The solution:

11. Escape velocity The velocity is positive for all t>0 or r>R: The escape velocity: The escape velocity from the earth

12. Escape velocity The escape velocity from the moon It is just over one-fifth of the escape velocity from the earth’s surface!!! The escape velocity: –The heavier the sphere, the larger the velocity –The thicker the sphere, the smaller the velocity

13. Models for pendulum motion Without damping With damping