Oscillations and Waves

Mr. Klapholz Shaker Heights High School Oscillations and Waves Problem Solving

Problem 1 A mass is hanging from a spring. It vibrates with a frequency of 0.5000 Hz. When t=0, the mass is released from a position that is 10.00 cm above equilibrium. What is the displacement of the pendulum after 1.000 second, 1.500 seconds, and 2.000 seconds?

Solution 1, part 1 The position of the mass is given by: x = Acos(wt) For this system, A = 10.00 cm, and the angular frequency is: • = 2pf = 2p(0.5000 s-1) = 3.142 Hz So, x = (10.0 cm) cos(3.142 t )

Solution 1, part 2 x = (10.0 cm) cos(3.142 t ) t = 1.000 sx = -10.00 cm t = 1.500 sx = 0.000 cm t = 2.000 sx = +10.00 cm Since the argument of the cosine must be an angle, the frequency that goes in the cosine is crafted so that after one period in the physical world, the argument has the value of 2p. We call it the angular frequency.

Problem 2 For the same pendulum as in Problem 1, what is the acceleration of the pendulum as when the displacement is 6.000 cm?

Solution 2 a = -w2x a = -(3.142 s-1)2(6.000 cm) a = -(3.142 s-1)2(6.000 cm) a = -59.22 cm s-2 a = -0.5922 m s-2

Problem 3 For the same pendulum of the previous two problems, what is the maximum velocity of the pendulum? Also, where is the pendulum when it has its maximum velocity?

Solution 3 The pendulum has its maximum velocity at the bottom of the motion: x = 0. The size of the velocity can be calculated from velocity = -Awsin(wt) vmax = Aw = (10.00 cm)(3.142 Hz) = 31.42cm s-1 Or v2 = w2 (A2 – x2)  v2 = (3.142)2 (102 – 02) v = 31.42cm/s

Problem 4 a) A wooden cylinder is floating in water, with it’s flat sides at the top and bottom. If you give it a push, so that more of it is submerged, then it will accelerate: a = -(rg / sL) x . L = the length of the cylinder. • = the density of water. • = the density of wood. x = the displacement of the cylinder from equilibrium Explain why the cylinder will display simple harmonic motion.

Solution 4 a) Since the acceleration is opposite (and proportional) to the displacement, we know that the motion will be simple and harmonic.

Problem 4 b) The cylinder is 52 cm long. The initial displacement is 24 cm. The density of water is 1000 kg per cubic meter. The density of wood is 840 kg per cubic meter. What is the maximum acceleration of the wood?

Solution 4 b) a = -{ rg / sL } x The greatest acceleration will happen when the displacement is its greatest. a = -{1000×9.8 / (840×0.24) } 0.24. a = -5.4 m s-2 The greatest acceleration will be a = 5.4 m s-2.

Problem 5 Water waves on the ocean have crests that are separated by 12 meters. The waves splash against a boat every 4.0 seconds. What is the speed of the waves?

Solution 5 F = 1 / T = 1 / 4.0 s = 0.25 s-1 = 0.25 Hz v = fl = (0.25 Hz) (12 m) = 3 m s-1.

Problem 6 In the graph that follows, Find the • Period • Amplitude • Maximum speed • Speed at 17 ms • Maximum acceleration. • Times when the velocity is zero. • Times when the acceleration is zero.

Use this graph in Problem 6 http://www.allaboutcircuits.com/worksheets/wave.html

Solution 6 a) Period = 20 ms = 0.02 seconds b) Amplitude = 6 V (we don’t know this unit yet) c) Maximum speed. Recall that v = -Awsin(wt). So the maximum speed is Aw. • = 2p/T = 2p/0.02 s = 314 Hz. Max Speed = Aw = (6)(314) = 1885 ≈ 2000

Solution 6 d) Speed at 17 ms v = -Awsin(wt) = -(6)(314)sin( 314×0.017 ) Use your calculator in radians (convert to degrees). v = 1527 v ≈ 2000 e) Maximum acceleration. Recall that a = -Aw2cos(wt). Max Acceleration= Aw2 = (6)(314) 2 ≈ 600,000

Solution 6 f) The velocity is zero when the position is a maximum or a minimum. That when the object is changing direction: 5 ms, 15 ms, 25 ms, etc. (Remember that each division is worth 2 ms) g) Times when the acceleration is zero. The acceleration is zero when the object has no force on it; that’s when the object is moving through equilibrium: 10 ms, 20 ms, 30 ms, …

Problem 7 Water waves are in a deep region and move at 20 m s-1. The waves then reach a more shallow region and travel at 15 m s-1. The original angle was 50˚; what is the angle after the change in direction? What is the new frequency? Please see the diagram:

http://www.explorelearning.com/index.cfm?method=cResource.dspExpGuide&ResourceID=552http://www.explorelearning.com/index.cfm?method=cResource.dspExpGuide&ResourceID=552

Solution 7 v2 sin(q1) = v1 sin(q2) 15sin(50˚) = 20sin(q2) 15 × 0.766 = 20sin(q2) 11.49 = 20sin(q2) 0.575 = sin(q2) q2 = InvSin( 0.575 ) q2 = 35.1˚ The direction, the speed, and the wavelength all change. The frequency stays the same.

Problem 8 A mass is on a frictionless surface, oscillating due to a spring. The spring has a constant of 74 N m-1, and it gets compressed and stretched. The amplitude of the motion is 8.0 cm. a) What is the total energy of the system? b) When the displacement is 4.8 cm, how much potential energy, and how much kinetic energy are in the system?

Solution 8 a) The total energy stays constant. At the maximum displacement, the total energy is all elastic. Elastic energy = (½)kx2 Elastic energy = (½)(74 N/m)(0.080 m)2 = 0.24 J b) When x = 0.048 m, then the Elastic energy = (½)(74)(0.048)2 = 0.085 J. Total Energy = Kinetic + Elastic Potential So the kinetic energy is 0.24 – 0.085 = 0.16 J

Problem 9 There is a type of glass in which light travels at 2.0 x 108m s-1. What is the index of refraction of this glass?

Solution 9 n = c / v n = { 3.0 x 108m s-1 } / {2.0 x 108m s-1 } n = 1.5 The index of refraction is a ratio of speeds; it has no units.

Tonight’s HW: Go through the Oscillations and Waves section in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.

Oscillations and Waves