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Chapter 13 Solutions - Mosinee School District

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<strong>Chapter</strong> <strong>13</strong><br />

f<br />

v boat wave<br />

, 5.0 m s<br />

20 m<br />

0.25 Hz<br />

<strong>13</strong>.49 The down and back distance is 4.00 m + 4.00 m = 8.00 m.<br />

The speed is then<br />

v<br />

d total<br />

t<br />

4 8.00 m<br />

0.800 s<br />

40.0 m s<br />

F<br />

Now,<br />

m<br />

L<br />

0.200 kg<br />

4.00 m<br />

5.00 10<br />

2<br />

kg m<br />

so<br />

F v<br />

5.00 10 kg m 40.0 m s 80.0 N<br />

2 2<br />

2<br />

<strong>13</strong>.50 The speed of the wave is<br />

v<br />

x<br />

t<br />

20.0 m<br />

0.800 s<br />

25.0 m s<br />

and the mass per unit length of the rope is m/ L 0.350 kg m . Thus, from v F , we obtain<br />

2<br />

2<br />

25.0 m s 0.350 kg m 219 N<br />

F v<br />

<strong>13</strong>.51 (a) The speed of transverse waves in the cord is v F , where mL is the mass per unit length. With<br />

the tension in the cord being F = 12.0 N, the wave speed is<br />

v<br />

F F FL<br />

m L m<br />

12.0 N 6.30 m<br />

0.150 kg<br />

22.4 m s<br />

(b)<br />

The time to travel the length of the cord is<br />

t<br />

L<br />

v<br />

6.30 m<br />

22.4 m s<br />

0.281 s<br />

<strong>13</strong>.52 (a) In making 6 round trips, the pulse travels the length of the line 12 times for a total distance of 144 m. The<br />

speed of the pulse is then<br />

Page <strong>13</strong>.20

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