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Unformatted text preview: P, oscillates with simple harmonic motion in the vertical direction. If the wave at t
can be written as 0 is as described in Figure 16.19b, then the wave function
A sin(kx y t) We can use this expression to describe the motion of any point on the string. The
point P (or any other point on the string) moves only vertically, and so its x coordinate remains constant. Therefore, the transverse speed vy (not to be confused
with the wave speed v ) and the transverse acceleration ay are
vy
ay dy
dt
dv y
dt constant x x y
t constant A cos(kx
vy
t 2A sin(kx (16.16) t)
t) (16.17) In these expressions, we must use partial derivatives (see Section 8.6) because y depends on both x and t. In the operation y/ t, for example, we take a derivative
with respect to t while holding x constant. The maximum values of the transverse
speed and transverse acceleration are simply the absolute values of the coefﬁcients
of the cosine and sine functions:
v y, max A (16.18) a y, max 2A (16.19) The transverse speed and transverse acceleration do not reach their maximum values simultaneously. The transverse speed reaches its maximum value ( A ) when
y 0, whereas the transverse acceleration reaches its maximum value ( 2A ) when
y
A. Finally, Equations 16.18 and 16.19 are identical in mathematical form to
the corresponding equations for simple harmonic motion, Equations 13.10 and
13.11. 507 16.8 Rate of Energy Transfer by Sinusoidal Waves on Strings Quick Quiz 16.4
A sinusoidal wave is moving on a string. If you increase the frequency f of the wave, how do
the transverse speed, wave speed, and wavelength change? EXAMPLE 16.4 A Sinusoidally Driven String The string shown in Figure 16.19 is driven at a frequency of
5.00 Hz. The amplitude of the motion is 12.0 cm, and the
wave speed is 20.0 m/s. Determine the angular frequency
and angular wave number k for this wave, and write an expression for the wave function. Solution Using Equations 16.10, 16.12, and 16.13, we ﬁnd Because A
y 12.0 cm A sin(kx 0.120 m, we have
t) (0.120 m) sin(1.57x 31.4t ) Exercise Calculate the maximum values for the transverse
speed and transverse acceleration of any point on the string. that
2
T
k 16.8 v 2f 2 (5.00 Hz) 31.4 rad/s
20.0 m/s 31.4 rad/s Answer 3.77 m/s; 118 m/s2. 1.57 rad/m R ATE OF ENERGY TRANSFER BY SINUSOIDAL
WAVES ON STRINGS As waves propagate through a medium, they transport energy. We can easily
demonstrate this by hanging an object on a stretched string and then sending a
pulse down the string, as shown in Figure 16.20. When the pulse meets the suspended object, the object is momentarily displaced, as illustrated in Figure 16.20b.
In the process, energy is transferred to the object because work must be done for
it to move upward. This section examines the rate at which energy is transported
along a string. We shall assume a onedimensional sinusoidal wave in the calculation of the energy transferred.
Consider a sinusoidal wave traveling on a string (Fig. 16.21). The source of the
energy being transported by the wave is some external agent at the left end of the
string; this agent does work in producing the oscillations. As the external agent
performs work on the string, moving it up and down, energy enters the system of
the string and propagates along its length. Let us focus our attention on a segment
of the string of length x and mass m. Each such segment moves vertically with
simple harmonic motion. Furthermore, all segments have the same angular frequency and the same amplitude A. As we found in Chapter 13, the elastic potential energy U associated with a particle in simple harmonic motion is U 1ky 2,
2
where the simple harmonic motion is in the y direction. Using the relationship
2
k /m developed in Equations 13.16 and 13.17, we can write this as ∆m Figure 16.21 A sinusoidal wave
traveling along the x axis on a
stretched string. Every segment
moves vertically, and every segment
has the same total energy. m
(a) m
(b) Figure 16.20 (a) A pulse traveling to the right on a stretched
string on which an object has been
suspended. (b) Energy is transmitted to the suspended object when
the pulse arrives. 508 CHAPTER 16 Wave Motion U 1m 2y 2. If we apply this equation to the segment of mass m, we see that the
2
potential energy of this segment is
1
2( U 2y 2 m) Because the mass per unit length of the string is
potential energy of the segment as
1
2( U m/ x, we can express the 2y 2 x) x : dx, and this expression be As the length of the segment shrinks to zero,
comes a differential relationship:
1
2( dU 2y 2 dx) We replace the general displacement y of the segment with the wave function for a
sinusoidal wave:
dU 1
2 2[A 1
2 t )]2 dx sin(kx 1
2 sin2(kx t ) dx 0, then the potential energy in a given If we take a snapshot of the wave at time t
segment is
dU 2A2 2A2 sin2 kx dx To obtain the total potential energy in one wavelength, we integrate this expression over all the string segments in one wavelength:
U dU
0
1
2 1
2 2A2 2A2 1 x
2 1
4k 1
2 sin2 kx dx sin 2 kx 1
2 0 2A2 sin2 kx dx
0 2A2(1
2 ) 1
4 2A2 Because it is in motion, each segment of the string also has kinetic en...
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This note was uploaded on 03/24/2010 for the course PHYSICS 2202 taught by Professor Mihalisin during the Spring '09 term at Temple.
 Spring '09
 MIHALISIN
 Physics

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