Chapter 13 Solutions - Mosinee School District
Chapter 13 Solutions - Mosinee School District
Chapter 13 Solutions - Mosinee School District
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<strong>Chapter</strong> <strong>13</strong><br />
and from Equation <strong>13</strong>.14c, the acceleration at this time is<br />
a A<br />
2<br />
cos t A k m cos t k m<br />
250 N m 250 N m<br />
5.00 10 m cos 0.500 s 4.59 m s<br />
0.500 kg 0.500 kg<br />
-2 2<br />
<strong>13</strong>.33 From Equation <strong>13</strong>.6,<br />
k<br />
v A x A x<br />
m<br />
2 2 2 2 2<br />
Hence,<br />
v A2 A2 cos2 t A 1 cos2<br />
t A sin t<br />
From Equation <strong>13</strong>.2,<br />
k<br />
m<br />
a x<br />
2<br />
A cos t<br />
2A cos t<br />
<strong>13</strong>.34 (a) The height of the tower is almost the same as the length of the pendulum. From T 2<br />
L g , we obtain<br />
L<br />
9.80 m s2<br />
15.5 s<br />
gT2<br />
4 4<br />
2 2<br />
2<br />
59.6 m<br />
(b)<br />
On the Moon, where g = 1.67 m/s 2 , the period will be<br />
T<br />
L 59.6 m<br />
2 2 37.5 s<br />
g 1.67 m s2<br />
<strong>13</strong>.35 The period of a pendulum is the time for one complete oscillation and is given by T 2<br />
g , where l is the<br />
length of the pendulum.<br />
(a)<br />
T<br />
3.00 min 60 s<br />
120 oscillations 1 min<br />
1.50 s<br />
(b)<br />
The length of the pendulum is<br />
g<br />
T<br />
2<br />
1.50 s<br />
9.80 m s2<br />
0.559 m<br />
4<br />
2<br />
4<br />
2<br />
2<br />
Page <strong>13</strong>.14
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