The simple pendulum

# The simple pendulum

## The simple pendulum

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1. The simple pendulum θ L m

2. The simple pendulum θ L m mg

3. The simple pendulum θ L m mg sinθ mg

4. The simple pendulum θ L x m mg sinθ mg

5. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L x m mg sinθ mg

6. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L + x m mg sinθ mg

7. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ + x m mg sinθ mg

8. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) + x m mg sinθ mg

9. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L + x m mg sinθ mg

10. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma + x m mg sinθ mg

11. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma ma = - mg x L + x m mg sinθ mg

12. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma ma = - mg x L + x m mg sinθ mg

13. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma and a = - g x L + x m mg sinθ mg

14. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma and a = - g x L Compare with SHM equation: a = - (2πf)2 x + x m mg sinθ mg

15. and a = - g x L Compare with SHM equation: a = - (2πf)2 x θ L + x m mg sinθ mg

16. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L θ L + x m mg sinθ mg

17. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L θ L + x m mg sinθ mg

18. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L T = 2π L g θ L + x m mg sinθ mg

19. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L T = 2π L g θ L + x Discuss: effect of length, mass, gravity, angle of swing. m mg sinθ mg

20. T = 2πL g Put in the form: y = m x + c

21. T = 2πL g Put in the form: y = m x + c T 2 = 4 π 2L + 0 g

22. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m

23. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mg

24. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mv2 = Ts - mg r mg

25. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mv2 = Ts - mg but r = L so mv2 = Ts - mg r L mg