Problem 12B

NAME ______________________________________ DATE _______________ CLASS ____________________Holt PhysicsProblem 12BPERIOD OF A SIMPLE PENDULUMPROBLEMSOLUTIONTwo friends in France use a pendulum hanging from the world’s highestrailroad bridge to exchange messages across a river. One friend attaches aletter to the end of the pendulum and releases it so that the pendulumswings across the river to the other friend. The bridge is 130.0 m abovethe river. How much time is needed for the letter to make one swingacross the river? Assume the river is 16.0 m wide.Given: L = 130.0 m g = 9.81 m/s 2Unknown: t = time required for pendulum to cross river = T/2 =?Use the equation for the period of a simple pendulum. Then divide the period bytwo to find the time of one swing across the river. The width of the river is notneeded to calculate the answer, but it must be small compared to the length ofthe pendulum in order to use the equations for simple harmonic motion.T = 2p ⎯ L g ⎯ = 2p ⎯ 19 . 8 31 0. 0⎯ m m / s 2 = 22.9 st = ⎯ T 2 ⎯ = ⎯22 .9 s⎯ =211.4 sADDITIONAL PRACTICECopyright © by Holt, Rinehart and Winston. All rights reserved.1. An earthworm found in Africa was 6.7 m long. If this worm were a simplependulum, what would its period be?2. The shortest venomous snake, the spotted dwarf adder, has an averagelength of 20.0 cm. Suppose this snake hangs by its tail from a branch andholds a heavy prey with its jaws, simulating a pendulum with a length of15.0 cm. How long will it take the snake to swing through one period?3. If bamboo, which can grow 88 cm in a day, is grown for four days and thenused to make a simple pendulum, what will be the pendulum’s period?4. A simple pendulum with a frequency of 6.4 × 10 −2 Hz is as long as thelargest known specimen of Pacific giant seaweed. What is this length?5. The deepest permafrost is found in Siberia, Russia. Suppose a shaft isdrilled to the bottom of the frozen layer, and a simple pendulum with alength equal to the depth of the shaft oscillates within the shaft. In 1.00 hthe pendulum makes 48 oscillations. Find the depth of the permafrost.6. Ganymede, the largest of Jupiter’s moons, is also the largest satellite inthe solar system. Find the acceleration of gravity on Ganymede if a simplependulum with a length of 1.00 m has a period of 10.5 s.Problem 12B 121

Givens7. h = 1.02 × 10 4 mL = 4.20 × 10 3 mk = 3.20 × 10 −2 N/mSolutionsF = kx = k(h − L)F = (3.20 × 10 −2 N/m)(6.0 × 10 3 m) = 190 N8. h = 348 mL = 2.00 × 10 2 mk = 25.0 N/mg = 9.81 m/s 2F = kx = k(h − L) = mgm =(25.0 N/m)(148 m)⎯⎯(9.81 m/s 2 )= 377 kgAdditional Practice 12B1. L = 6.7 mg = 9.81 m/s 2 T = 2π ⎯ L g ⎯ = 2π ⎯ (9 ( .86.7m)⎯ 1 m / s2 =)5.2 sII2. L = 0.150 m(0.150 m)⎯⎯(9.81 m/s 2 )g = 9.81 m/s 2 T = 2p ⎯L g ⎯ ⎯ = 2p = 0.777 s3. x = 0.88 mg = 9.81 m/s 2T = 2p ⎯ 4 x⎯ =g4(0.88 m)2p ⎯⎯(9.81 m/s 2 )T = 3.8 s4. f = 6.4 × 10 −2 Hzg = 9.81 m/s 2 T = ⎯ 1 f ⎯⎯ = 2p ⎯L g ⎯ ⎯ g (9.81 m/s 2 )L = ⎯ 4p 2 ⎯f 2 = ⎯⎯⎯4π 2 (6.4 × 10 −2 Hz) 25. t = 3.6 × 10 3 sN = 48 oscillationsL = 61 mg = 9.81 m/s 2 T = 2p ⎯ L g ⎯ ⎯ = ⎯ N6. L = 1.00 mT = 10.5 st ⎯ ⎯ tN⎯2g(3.6 × 10 3 s) 2 (9.81 m/s 2 )L = ⎯ = ⎯⎯⎯4p 24p 2 (48) 2L = 1.4 × 10 3 mg = ⎯ 4p 2L 4p 2 (1.00 m)⎯T2= ⎯⎯ = 0.358 m/s 2(10.5 s) 2Copyright © by Holt, Rinehart and Winston. All rights reserved.II Ch. 12–2Holt Physics Solution Manual