course Phy232
Assignment 11 Ch. 15 Problems: 47, 48, 55, 59, 6447. (A): What are the frequency, angular frequency, and wave number?
Wave #: 2pi/wavelength= 2pi/1.80=3.49
Frequency: v=wavelength x frequencies = 36.0=1.80f= 36/1.8= 20
Angular Frequency: 126rad/s(t) w=2piF=2pi(20)=125.6
(B): What is the function y(x,t) that describes the wave?
Y(x,t): (2.50mm) x cos[(3.49rad/mx)x-(126rad/s)t
(c): What is y(t) for a particle at the left end of the string?
Y(t)=(2.50mm)cos[(126rad/s)t]
(D): What is y(f) for a particle 1.35m to right of the origin?
Y(t): (2.50mm)x cos [3pi/2 = (126 rad/s)t]
(E): What is maximum magnitude of transverse velocity of any point?
Not sure
If the point is specified by its coordinate x, its transverse velocity at any instant is dy/dt, evaluated at that instant and for the x coordinate.
(F): Transverse velocity and displacement particle 1.35m to right at t=0.0625s
Y(x,t)=(2.50mm)x cos (3.49)x-(126)t
= -2.50
48. y(x,t)=(0.750cm)cos pi[(0.400cm)x + (250s)t]
(A): Find amplitude, period, frequency, wavelength, and speed.
Amplitude: 0.750cm
Frequency: w=2pif
250=2pif=39.8hz
Wavelength: wave#=2pi/wavelength 0.400=2pi/wavelength =2.72cm
0.400=2pi/wavelength =2.72cm is in the first place a false statement, in that it implies that .400 = 2.72 cm. Avoid misuse of the equal sign, which indicates equality, not train of thought.
If you correctly solve the equation .400 = 2 pi / wavelength for wavelength you get wavelength = 2 pi / (.400 m^-1) = 5 pi m, or about 16 m.
However in this case the wave number is .400 pi so wavelength is 2 pi / (.400 m^-1 pi) = 5 m.
The frequency is 2 pi / (
Speed= fxwavelength = 108.3cm/s
Period: T=1/f = 1/39.8 =0.025s
(B): Sketch the shape of rope at values t=0,0.0005s and 0.0010
plug a t value into the equation of the wave, and you get a function of x. Graph that function to see the shape and position of the wave at that instant.
(C): tx direction
(D): mass/length of rope= 0.0500kg/m Find the Tension
V^2= T/(m/l)
V^2 x (m/l) =T
T= )6.25 m/s)^2(0.5)
T= 19.53N
(E): Find average power?
Pav=1/2sqrt (m/l)F x omega^2 x A^2
Pav= ½ sqrt 0.27 x 19.5 x (108.3)^2(0.0075)^2
=1/2sqrt(5.265) x (11728.89)(5.625e-5)
=½(2.29) x 0.6597
= 1.145 x 0.6597
= 0.76 watts
55. Wave propagating to right in tx direction
(A): Determine direction of transverse velocity of each 6 labeled points
1=0, 2=+, 3= -, 4= 0, 5= -, 6= +
(B): Determine the direction of transverse acceleration
1=-, 2=+, 3= -, 4= +, 5=-, 6=0
(C): In “a” it would reverse the sign but in “b” there would be no charge.
you didn't include enough information about the problem for me to verify your solutions.
59. Waves of Arbitrary Shape
(A): It displaces some distance y in the direction perpendicular to x-axis.
(B): y(x,t)= f(u) u=x-vt
Y(xt/x=df(u)/du=df(u)/du
(C); Wave Pulse= y(x,t)De^-(Bx-Ct)^2
C/B
I can't tell everything from the info you give.
f(x - v t) can be seen as f(x) shifted v t units in the x direction; so as t increases the shift increases, and the wave moves in the x direction at velocity v.
&#Good responses. Let me know if you have questions. &#
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