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course PHY 232
3/15 3:25 pm
110221 Physics`q001. A standing wave in a string has three antinodes between its ends, and the ends are nodes. The ends are separated by 6 meters.
What is wavelength of the wave?
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There are 1.5 wavelengths in the string since 1 wavelength has NANAN pattern and this pattern is NANANAN which is an extra AN over 1 wl.
wl = 6*2/3 = 4 m
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If the propagation velocity in the string is 30 meters / second, what is the frequency of the wave (i.e., how many complete cycles occur per second)?
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v= lamba*f
f = v/lamba = 30 m/s / 4 m/cyle = 7.5 cycles/s = 7.5 Hz
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`q002. A standing sound wave in a pipe 4 meters long has an antinode at one end and a node at the other. Nodes and antinodes are equally spaced, and two antinodes occur within the pipe. Sound waves propagate at about 340 meters / second. What is the frequency of this wave?
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The pattern is NANANA, which is 1/4 greater than 1 cycle so
4/5 = 0.8 --> 4*0.8 = 3.2 m
f = (340 m/s) / (3.2 m/cycle) = 106.25 cycles/s = 106.25 Hz
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`q003. A string has nodes at both of its ends, which are 3.6 meters apart. It is theoretically possible to achieve a standing wave of any wavelength, provided that nodes and antinodes alternate at equal spacing.
What therefore are the six longest wavelengths that can occur in this string?
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Each successive AN addition adds 1/4 wavelength.
1: NAN--> 4/2*3.6 = 7.2 m
2: NANAN--> 4/4*3.6 = 3.6 m
3: NANANAN --> 4/6*3.6 = 2.4 m
4: 1.8 m
5: 1.44 m
6: 1.2 m
Generally:
lambda_n =lamda_1/(2n)
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If the first of these has a frequency of 30 cycles / second, what is the speed of propagation in the string?
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v = 30*7.2 = 216 m/s
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What then are the frequencies of the other five wavelengths?
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f = v/lambda
2: 60
3: 90
4: 120
5: 150
6: 180
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`q004. A sound wave in a pipe has a node at one end and an antinode at the other. The frequency of the longest possible wave in this pipe is 20 Hz.
What are the three next-lowest frequencies possible in this pipe? Once more, nodes and antinodes must alternate and must be equally spaced.
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So the wavelength gets shorter meaning f gets larger. In fact it's the reciprocal of the wavelength change.
f_1= 3*20 = 60 Hz
f_2 = 5*20= 100
f_3 = 7*20 = 140 Hz
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How long is the pipe? Recall that sound in air travels at about 340 m/s.
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lambda_0 = 340/20 = 17 m
NA=1/4 lambda--> L= 17/4 = 4.25 m
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University Physics:
`q005. The superposition of traveling waves y_1 ( x, t ) = A/2 * sin( k x - omega t) and y_2(x, t) = A / 2 * sin(k x + omega t) yields the equation y(x, t) = A sin(omega t) cos(k x), as shown in the pictures below.
Sketch the wave as it appears at clock time t = pi / (2 omega), between x = 0 and x = 6 pi / k.
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Here:
sin (omega*(π/(2omega)) = 1
The cos oscillates between A/-A 6 times in the interval.
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Sketch between the same to x values, for clock time t = 2 * pi / (2 omega), then again at t = 3 pi / (2 omega).
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The first t makes sin = 0 and so all values are 0.
The second t inverts the graph above so that it goes -A->0->A instead of A->0->-A
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What will the sketch look like at t = 4 pi / (2 omega)?
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here sin= 0 so all y =0
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Does the standing wave have nodes at x = 0 and x = 6 pi / k?
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In order to be a node it is required that y=0 at all values of t which only happens when cos = 0 or kx is some multiple of π/2.
Here cos 0 & cos 6π = 1 which means y does change as t changes therefore these are not nodes.
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If k = pi/2 rad / meter and omega = 6 pi rad / sec, what are the wavelength and frequency of the traveling waves that combined to form this standing wave?
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lamba = 2π/k = 2π/(π/2)= 4 m
f = omega/2π= 6π/2π = 3 Hz
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What is the propagation speed of those traveling waves?
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c = omega/k = (6π)/(π/2) = 12 m/s
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How far is it from x = 0 to x = 6 pi / k?
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x= 6π/(π/2) = 12 m
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How many nodes does the standing wave have between its ends?
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6
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How many node-antinode distances are there?
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12
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Use the number of node-antinode distances to predict the wavelength.
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N-A = 1/4 lambda --> 12*1/4 = 3 thus 3 wavelengths in 12m so lambda = 4 m
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`q006. What is the equation of the simple harmonic motion of the particle whose position is x?
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y = A cos kx sin omega*t
where cos kx is constant and determines how close A pt x ever comes or is 0
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What is the maximum speed of the transverse motion of the particle whose position is x?
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dy/dt = A cos kx * omega* cos omega*t
dy/dt max @ cos omega*t = 1 so max dy/dt = A*omega cos kx
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If the mass density of the medium is m / L, then what is the approximate maximum KE of a short segment of length `dx, which includes the particle at position x?
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KE_max @ v_max and m= m/L* `d x
KE_max_dx= 1/2 (m/L)*dx * (A*omega cos kx)^2
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This should suggest an integral for the total energy in the wave, between position x = 0 and position x = 6 pi / k. What is the integral? What do you get when you do the integration?
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The expression is:
1/2 (m/L)*dx * (A*omega cos kx)^2
Integrating the above expression gives:
KE_max = (A*omega)^2*1/2m/L*[(sin 2kx)/4k+x/2]
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@& Really well done.*@