16 - Wave Motion

Figure 1611a shows such a segment at the instant it

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the pulse, and then continues to move to the left. Figure 16.11a shows such a segment at the instant it is located at the top of the pulse. The small segment of the string of length s shown in Figure 16.11a, and magnified in Figure 16.11b, forms an approximate arc of a circle of radius R . In our moving frame of reference (which is moving to the right at a speed v along with the pulse), the shaded segment is moving to the left with a speed v. This segment has a centripetal acceleration equal to v 2/R, which is supplied by components of the tension T in the string. The force T acts on either side of the segment and tangent to the arc, as shown in Figure 16.11b. The horizontal components of T cancel, and each vertical component T sin acts radially toward the center of the arc. Hence, the total radial force is 2 T sin . Because the segment is small, is small, and we can use the small-angle approximation sin . Therefore, the total radial force is Fr 2T sin 2T s. Because the segment forms part of a circle The segment has a mass m at the center, s R(2 ), and hence and subtends an angle 2 m s 2R 501 16.5 The Speed of Waves on Strings If we apply Newton’s second law to this segment, the radial component of motion gives Fr 2T ma mv 2 R 2 R v2 R Solving for v gives Equation 16.4. Notice that this derivation is based on the assumption that the pulse height is small relative to the length of the string. Using this assumption, we were able to . Furthermore, the model assumes that the tenuse the approximation sin sion T is not affected by the presence of the pulse; thus, T is the same at all points on the string. Finally, this proof does not assume any particular shape for the pulse. Therefore, we conclude that a pulse of any shape travels along the string with speed v √T/ without any change in pulse shape. EXAMPLE 16.2 The Speed of a Pulse on a Cord A uniform cord has a mass of 0.300 kg and a length of 6.00 m (Fig. 16.12). The cord passes over a pulley and supports a 2.00kg object. Find the speed of a pulse traveling along this cord. Solution The tension T in the cord is equal to the weight of the suspended 2.00-kg mass: T (2.00 kg)(9.80 m/s2) mg 19.6 N (This calculation of the tension neglects the small mass of the cord. Strictly speaking, the cord can never be exactly horizontal, and therefore the tension is not uniform.) The mass per unit length of the cord is m 5.00 m 0.300 kg 6.00 m 0.050 0 kg/m Therefore, the wave speed is 1.00 m 2.00 kg Figure 16.12 The tension T in the cord is maintained by the suspended object. The speed of any wave traveling along the cord is given by v √T/ . v √√ T Exercise 19.6 N 0.050 0 kg/m 19.8 m/s Find the time it takes the pulse to travel from the wall to the pulley. Answer 0.253 s. Quick Quiz 16.3 Suppose you create a pulse by moving the free end of a taut string up and down once with your hand. The string is attached at its other end to a distant wall. The pulse reaches the wall in a time t. Which of the following actions, taken by itself, decreases the time it takes the pulse to reach the wall? More than one choice may be correct. (a) Moving your hand more quickly, but still only up and down once by the same amount. (b) Moving your hand more slowly, but still only up and down once by the same amount. (c) Moving your hand a greater distance up and down in the same amount of time. (d) Moving your hand a lesser distance up and down in the same amount of time. (e) Using a heavier string of the same length and under the same tension. (f) Using a lighter string of the same length and under the same tension. (g) Using a string of the same linear mass density but under decreased tension. (h) Using a string of the same linear mass density but under increased tension. 502 CHAPTER 16 16.6 Incident pulse (a) (b) (c) (d) (e) Reflected pulse Figure 16.13 The reflection of a traveling wave pulse at the fixed end of a stretched string. The reflected pulse is inverted, but its shape is unchanged. Incident pulse (a) Wave Motion REFLECTION AND TRANSMISSION We have discussed traveling waves moving through a uniform medium. We now consider how a traveling wave is affected when it encounters a change in the medium. For example, consider a pulse traveling on a string that is rigidly attached to a support at one end (Fig. 16.13). When the pulse reaches the support, a severe change in the medium occurs — the string ends. The result of this change is that the wave undergoes reflection — that is, the pulse moves back along the string in the opposite direction. Note that the reflected pulse is inverted. This inversion can be explained as follows: When the pulse reaches the fixed end of the string, the string produces an upward force on the support. By Newton’s third law, the support must exert an equal and opposite (downward) reaction force on the string. This downward force causes the pulse to invert upon reflection. Now consider another case: this time, the pulse arrives at the end of a...
View Full Document

This note was uploaded on 03/24/2010 for the course PHYSICS 2202 taught by Professor Mihalisin during the Spring '09 term at Temple.

Ask a homework question - tutors are online