King Fahd University of Petroleum and Minerals – Physics Department Physics 102 Recitation – Term 172 – Spring 2018 – Section 7 Quiz # 1 Name: ID # _______ __ A transverse sinusoidal wave is travelling to the right (+ ݔ-axis) on a stretched string. The general form of the wave is ݕ(ݐ,ݔ)= ݕsin( ݔ± ݐ+ ). The amplitude of the wave is 0.800 cm, its wavelength is 0.550 m, and its phase constant is zero. The speed of waves on the string is 24.0 m/s. What is the speed of the particle of the string located at ݔ= 0.300 m when the time ݐis 1.50 s? SOLUTION: = 2 ߨ ߣ= =ݒ2 ߣ/ݒߨݕ(ݐ,ݔ)= ݕsin( ݔ−ݐ) ݑ(ݐ ,ݔ) = − ݕcos( ݔ−ݐ) ݑ(ݐ ,ݔ) = −(2ߣ/ݒߨ) ݕcosሾ(2ߣ/ݔߨ) − (2ߣ/ݐݒߨ)ሿ =ݑ∴− ൬2 ×ߨ24.0 0.550 ൰ × 8.00 × 10 ଷ cos ൬2 ×ߨ0.300 0.550 ൰ − ൬2 ×ߨ24.0 × 1.50 0.550 ൰൨ = −. ૡ ܛ/ܕ
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King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 1
Name: ID # _______ __
A transverse sinusoidal wave is travelling to the right (+ -axis) on a stretched
string. The general form of the wave is ( , ) = sin ( ± + ). The
amplitude of the wave is 0.800 cm, its wavelength is 0.550 m, and its phase
constant is zero. The speed of waves on the string is 24.0 m/s. What is the speed
of the particle of the string located at = 0.300 m when the time is 1.50 s?
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 1
Name: ID # _______ __
Two identical sinusoidal traveling waves, each with a frequency of 100 Hz, are moving
in the same direction along a stretched string. If they are interfering and the combined
wave has an amplitude that is 0.25 times that of the amplitude of each wave, calculate
the phase difference between the two waves in radians and in meters. The speed of
waves on the string is 25 m/s.
SOLUTION:
The amplitude of the resultant wave is : = 2 cos /2 .
Thus, the phase difference is : = 2 cos /2 = 2 cos 0.25 /2 = 2.89 rad
The wavelength is : = = = 0.25 m.
Therefore, the phase difference in meters is : ∆ = = 0.115 m.
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 1
Name: ID # _______ __
A standing wave is set up on a string of length 1.2 m that is fixed at its ends. The string vibrates
according to the equation: , = 0.50 sin 2.5 cos 40 , where and are in
meters and is in seconds.
(a) What is the speed of waves on this string?
(b) What is the harmonic number of the standing wave?
SOLUTION:
(a) = = . = 16 m/s.
(b) = ⟹ = = = = × . × . = 3
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 7
Quiz # 2
Name: ID #
The difference between two consecutive resonant frequencies in a tube closed at
one end is 80 Hz. What is the fifth harmonic frequency of this tube?
SOLUTION:
The resonant frequencies of the tube are given by: = , where n = 1, 3, 5, …
The difference between two consecutive resonant frequencies is: ∆ = − = − = .
The fifth harmonic frequency of the tube is = = = × 80 = 200 Hz.
DELL
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King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 2
Name: ID #
The intensity of sound is 3.20 × 10 W/m2 at a distance of 5.00 m from a
source. What is the sound level at a distance of 10.0 m from the source?
SOLUTION:
Recall that the intensity ( ) of sound at a distance from a source is: = . Let be the intensity at 5.00 m, and be the intensity at 10.0 m, then: = 4 × 100 ÷ 4 × 25 = 0.25
Thus, = 0.25 = 8.00 × 10 W/m .
Therefore, the sound level at a distance of 10.0 m is: = 10 log = 10 × log 8.00 × 10 10 = 89 dB
DELL
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King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 2
Name: ID #
A stationary speaker sends sound waves of frequency 1000 Hz toward a car
moving away at a speed of 35 m/s. What is the frequency of the waves reflected
back to the speaker? [speed of sound in air = 343 m/s]
The requested point is outside the shell; thus, the only charge that will contribute
to the electric field is the charge on the outer surface of the shell.
Since that charge is negative, the electric field is radially inward. = = × × . × . = / .
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 7
Name: ID #
Two infinite parallel lines of charge are shown in the figure below, and are separated
by 5.0 cm. Their uniform linear charge densities are = + 2.0 × 10 C/m and = − 3.0 × 10 C/m. Find the net electric field at a point that is 1.0 cm to the left
of . Express your answer in unit vector notation.
SOLUTION: The electric fields due to the lines are: = − ̂ = − × × × . × . × ̂ = − 3.6 × 10 ̂ (N/C) = + ̂ = + × × × . × . × ̂ = + 9.0 × 10 ̂ (N/C)
The net electric field is: = + = − 3.6 × 10 ̂ + 9.0 × 10 ̂ = − 2.7 × 10 ̂ (N/C)
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 9
Quiz # 7
Name: ID #
A cube of edge length 2.00 m is placed in a region in which the electric field is given
by = 20.0 − 28.0 (N/C). Determine the net charge contained within the cube.
SOLUTION: The area of each face is 4.0 m2.
Since the electric field is in the z direction, there will be electric flux only through
the top ( ) and bottom ( ) faces.
At the top (z = 2.0 m): Φ = . = − 36.0 . + 4.00 = − 144 N. m /C.
At the bottom (z = 0): Φ = . = + 20.0 . − 4.00 = − 80.0 N. m /C.
King Fahd University of Petroleum and Minerals – Physics Department
Physics 102 Recitation – Term 172 – Spring 2018 – Section 8
Quiz # 8
Name: ID #
Two isolated concentric conducting thin spherical shells have radii = 0.50 m and = 1.0 m; uniform charges = + 3.0 μC and = − 4.0 μC. With V = 0 at infinity,
what is the electric potential at a point that is 0.80 m from the center?