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56. Reference to point
A
as an antinode suggests that this is a standing wave pattern and
thus that the waves are traveling in opposite directions.
Thus, we expect one of them to
be of the form
y
=
y
m
sin(
kx
+
ω
t
) and the other to be of the form
y
=
y
m
sin(
kx
–
ω
t
).
(a) Using Eq. 1660, we conclude that
y
m
=
1
2
(9.0 mm) = 4.5 mm, due to the fact that the
amplitude of the standing wave is
1
2
(1.80 cm) = 0.90 cm = 9.0 mm.
(b) Since one full cycle of the wave (one wavelength) is 40 cm,
k
= 2
π/λ ≈
16 m
−
1
.
(c) The problem tells us that the time of half a full period of motion is 6.0 ms, so
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This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics

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