Question

The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 305 grams, and it is released at an angle of 11° to the vertical. (a) With what frequency does it vibrate? Assume SHM. Hz (b) What is the pendulum bob's speed when it passes through the lowest point of the swing? m/s (c) What is the total energy stored in this oscillation, assuming no losses? J

My Notes Ask Your Teacher 8. -13 points Giancoli6 11.P032 The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 305 grams, and it is released at an angle of 11° to the vertical. (a) With what frequency does it vibrate? Assume SHM (b) What is the pendulum bob's speed when it passes through the lowest point of the swing? (c) What is the total energy stored in this oscillation, assuming no losses? Hz m/s

Answer 1

(a)� T = L/f = squareroot 2 pi squareroot L/g rightwardsdoublearrow f = L/2 pi squareroot 2/L f = L/2 pi squareroot 9.81/0.66 rightwardsdoublearrow 0.613Hz (b) K epsilon lowest point = 1/2 mv^2 p epsilonHighest point = mgh energy conservation 1/2 mv^2 = mg delta h v^2 = 2gh = 2g(l � (cos 11^degree) v = squareroot 2 times 9.81 (0.66 � 0.66 cos (1) rightwardsdoublearrow 0.488m/s at lowest point (c) epsilon = 1/2mv^2 = Total energy (p epsilon = 0 lowest point) = 1/2 times (0.305) (0.488)^2 = 0.036J