1. A pendulum has a string of length L. To double the period, change the length to:
A. 2L
B. L/2
C. 4L.
D. L/4
Answer = .....
2. When particles in the medium vibrate at right angles to the direction of energy transport, then the wave is
A. longitudinal
B. standing
C. torsional
D. transverse
Answer = .....
3. As a wave travels along a uniform string, the wavelength of the pulse is
A. decreased
B. increased
C. unchanged
D. zero
Answer = .....
4. As a wave travels to a thicker string of the same material, the speed of the pulse is
A. decreased
B. increased
C. unchanged
D. zero
Answer = .....
5. An example for mechanical wave.
A. radio wave
B. light wave
C. infrared radiation
D. sound wave
Answer = .....
6. The intensity of a sound wave is
A. directly proportional to the distance from the source
B. inversely proportional to the distance from the source
C. directly proportional to the square of the distance from the source
D. inversely proportional to the square of the distance from the source
Answer = .....
7. What happens to a sound wave as it travels from air into water?
A. Increased intensity
B. Shorter Wavelength
C. Increased frequency
D. Unchanged frequency
Answer = .....
8. A sound wave is
A. a transverse wave
B. a longitudinal wave
C. an electromagnetic wave
D. torsional wave
Answer = .....
9. Once waves are no longer interacting with each other, the waves
A. trade energies
B. one wave gets consumed by the other
C. they double in size
D. the waves return to their original sizes
Answer = .....
10. According to the principle of superposition, displacement caused by combining two waves is equal to the
A. average amplitude of the contributing waves
B. amplitude of the larger wave
C. equilibrium position
D. sum of the amplitudes of contributing waves
Answer = .....
11. When two waves are in phase, ____________ interference will occur.
A. constructive
B. destructive
C. no
D. little
Answer = .....
12. Which type of wave DOES NOT requie a medium to propagate?
A. surface waves
B. transverse waves
C. electromagnetic waves
D. longitudinal waves
Answer = .....
13. The horizontal distance between successive crests of a wave is
A. frequency
B. amplitude
C. period
D. wavelength
Answer = .....
14. The highest point in a wave is
A. crest
B. trough
C. frequency
D. wavelength
Answer = .....
15. Sound travels at 340 ms^{1} in air and 5000 ms^{1} in rocks. The sound of an earthquake emerges into the air. Its
A. frequency remains the same but the wavelength is shorter.
B. frequency is higher but the wavelength remains the same.
C. frequency is lower but the wavelength remains the same.
D. frequency remains the same but the wavelength is longer.
Answer = .....
16. When an automobile moves towards a listener, the sound of its horn seems relatively
A. low pitched
B. high pitched
C. normal
D. No Change
Answer = .....
17. A trumpet player is standing on a stage playing a single note. Which choice would result in you hearing a higher pitch than the trumpeter?
A. Run very fast toward the trumpeter.
B. Cover your ears.
C. Play the sound much louder through a headphone.
D. Run very fast away from the trumpeter. .
Answer = .....
18. At a particular instant in time, a wave has its electric field pointing north and its magnetic field pointing up. In which direction is the wave traveling?
A. North
B. East
C. South
D. West
Answer = .....
19. All other variables being equal, which surface would experience the greatest the radiation pressure from sunlight?
A. black surface.
B. red surface.
C. yellow surface.
D. white surface.
Answer = .....
20. The image in a flat mirror is
A. real and upright
B. real and inverted
C. virtual and upright
D. virtual and inverted
Answer = .....
1. An oscillator consists of a block attached to a spring
(k = 400 Nm^{1}). At some time t, the position (measured from the system’s equilibrium location), velocity, and acceleration of the
block are x = 0.100 m, v = 13.6 ms^{1}, and a = 123 ms^{2}.
Calculate
(a) the frequency
of oscillation,
(b) the mass of the block, and
(c) the amplitude of the motion.
(a)ω^{2 }= (a_{max})/( x_{max})
ω = √[(123ms^{2})/( 0.100m)]
f = 5.58 Hz. = 5.58 Hz. (b) ω = √ [^{k}/_{m}]m= k/ω^{2}
m = [400Nm^{1}/rads s^{1}]
m = 0.325 kg. (c) By energy conservation E_{T} = E_{p} + E_{k}E_{T} = ½ kx^{2} + ½ mv^{2
}At the turning points where X is a maximum Ek = 0, so E_{T} = E_{p} and when x = 0 and V = v_{max} Ep = 0 so E_{T} = E_{k}.
E_{T} = ½ k(x_{max})^{2} + 0
E_{T} = ½ m (v_{max})^{2} + 0
We don’t know v_{max} but we know x and v.
½ k(x_{max})^{2} = ½ mv^{2} + ½ kx^{2}
k(x_{max})^{2} = (mv^{2}) + kx^{2
}(x_{max})^{2} = [(mv^{2})/k] + x^{2
}(x_{max})^{2} = {[(0.325kg)(13.6ms^{1})^{ 2}]/400Nm^{1}} + (0.100m)^{2} x_{max} = 0.400m
2. A wooden block of mass M = 5.4 kg, at rest on a horizontal
frictionless table,
is attached to a rigid support by a spring of constant 6000 Nm^{1}. A bullet of mass m = 9.5 g and horizontal velocity, v, 630 ms^{1} strikes and is embedded in the block.
Assuming the compression of the spring is negligible until the
bullet is embedded, determine (a) the
velocity
of the system immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.
(a) The collision is inelastic so kinetic energy is not conserved, but momentum is conserved in all collisions.
Momentum before collision = momentum after collision
MV + mv = (M+ m) v_{max }where v_{max }is the velocity of the block and bullet after the collision. As this velocity happens at the equilibrium position, it must be the maximum velocity with which the system will operate.
v_{max }= ^{[MV + mv]}/_{[M+ m]}
v_{max }= [(5.4 kg)(0 ms^{1}) + (0.0095 kg)(630ms^{1})] / [(5.4 kg) + (0.0095 kg)] v_{max }= 1.1 ms^{1}
(b) By energy conservation E_{T} = E_{p} + E_{k}E_{T} = ½ kx^{2} + ½ mv^{2
}At the turning points where X is a maximum Ek = 0, so E_{T} = E_{p} and when x = 0 and V = v_{max} Ep = 0 so E_{T} = E_{k}.
E_{T} = ½k(x_{max})^{2} + 0
E_{T} = ½(M+m) (v_{max})^{2} + 0
We know v_{max}.
½k(x_{max})^{2} = ½(M+m)(v_{max})^{2} (x_{max})^{2}= [(M+m)(v_{max})^{2}] / k
(x_{max})^{2} = [(5.4kg) + (9.5 X 10^{3} )] [(1.1 ms^{1})^{2}] / 6000Nm^{1} (x_{max}) = 3.3 X 10^{2} m
3. A nylon guitar string has a
linear density of 7.20 gm^{1} and is under
a tension of 150 N. The ﬁxed supports
are distance D, 90.0 cm apart. The
string is oscillating in a standing
wave pattern. Calculate the (a) speed, (b) wavelength,
and (c) frequency of the traveling waves whose superposition
gives this standing wave.
(a) The wave speed is given by v = √[τ/μ] where τ is the tension in the string and μ is
the linear mass density of the string.
Thus
v = √[(150N)/(7.20 X 10^{3}kgm^{1})] v = 1.44 X 10^{2} ms^{1}
(b) From the diagram, we find the wavelength of the standing wave to be λ = (2/3)(90.0 cm) λ = 0.60 m.
(c) The frequency is f = v/λ
f = (1.44 X 10^{2} ms^{1}) / (0.60 m) f = 240 Hz.
4. Earthquakes generate sound waves inside Earth.
Unlike a gas, Earth can experience both transverse (S) and longitudinal
(P) sound waves.
Typically, the speed of S waves is about 4.5
kms^{1}, and that of P waves 8.0 kms^{1}. A seismograph records P and S
waves from an earthquake. The ﬁrst P waves arrive 3.0 min before
the ﬁrst S waves. If the waves travel in a straight line, how far away did the earthquake occur?
(a) If
d is the distance from the location of the earthquake to the seismograph and v
is the
speed of the S waves, then the time for these waves to reach the seismograph is
t_{s} = [d/v_{s}] and similarly, the time for P waves to reach the seismograph is t_{p} = [d/v_{p}].
Time delay ΔT = t_{s}  t_{p}
ΔT = [d/v_{s}] [d/v_{p}]
ΔT = d(v_{p}v_{s}) / (v_{s}v_{p})
d = [(v_{s}v_{p}) ΔT] / (v_{p}v_{s})
d = {[(4.5 X 10^{3}ms^{1})(8.0 X 10^{3}ms^{1})(3.0 X 60s)] / [(8.0 X 10^{3}ms^{1}  (4.5 X 10^{3}ms^{1})]} d = 1.9 10^{6}m
5. Two
point sources S_{1} and S_{2} emit
sound of wavelength λ = 2.00 m.
The emissions are isotropic (the same intensity of radiation in all directions) and in
phase, and the separation between the sources, d is 16.0 m. At any
point P on the x axis, the wave fromS_{1} and the wave from S_{2 }interfere.
When P is very far away (x ≈ ∞),
what are (a) the phase difference
between the arriving waves from
S_{1} and S_{2} and (b) the type of interference they produce? Now
move point P along the x axis toward S_{1}. (c) Does the phase difference
between the waves increase or decrease? At
what distance x do the waves have a phase difference of (d) 0.50λ, (e)
1.00λ, and (f) 1.50λ?
(a) If point P is infinitely far away, then the small distance d between the two sources
is of no consequence (they seem effectively to be the same distance away from P). Thus,
there is no perceived phase difference.
(b) Since the sources oscillate in phase, then the situation described in part (a) produces
fully constructive interference.
(c) By Pythogoras, the distance S_{2}P is √[d^{2}+x^{2}]
So the path length difference for waves traveling from S_{1} and S_{2} ΔL = √[d^{2}+x^{2}]  x.
The phase difference in “cycles” (in absolute value) is Δφ = [ΔL / λ]
Δφ = {√[(d^{2}+x^{2})  x]} / λ
Δφ = {√[(d^{2}+x^{2})  x]} / λ
Consider ΔL = λ / 2 then √[d^{2}+x^{2}] becomes [x + λ / 2] (the diffrerence between the hypothenuse and x).
[d^{2}+x^{2}] = [x + λ / 2]^{2}
[d^{2}+x^{2}] = x^{2} + xλ + ( λ^{2} / 4)
x^{2}  x^{2}  xλ= d^{2}+ (λ^{2} / 4)
xλ= d^{2}  (λ^{2} / 4)
x= (d^{2} / λ)  (λ / 4)
So in general if ΔL = n λ where n > 0, then x= (d^{2} / 2 nλ)  (nλ / 2)
x= (d^{2} / 2nλ)  (nλ / 2)
x= [(16.0 m) ^{2} / 2n(2.00m)]  (n(2.00m) / 2)
x= {[(256) / 4n]  (n)}
x= {[(64) / n]  n} From this equation, it is clear that for d = 16.0 m and λ = 2.00 m that as x increases the phase difference
between the waves decreases.
(d) For ΔL = 0.50λ, x = [(64.0 / 0.50)  0.50] x = 127.5m
(d) For ΔL = 1.00λ, x = [(64.0 / 1.00)  1.00] x = 63.0m
(d) For ΔL = 1.50λ, x = [(64.0 / 1.50)  1.50] x = 41.2m
6.Suppose that the sound level of a conversation is initially at
an angry 70 dB and then drops to a soothing 50 dB. Assuming that
the frequency of the sound is 500 Hz, determine the (a) initial and
(b) ﬁnal sound intensities.
(a) β_{1} = (10 dB) log(I_{1} / I_{0})
70 dB = (10 dB) log(I_{1} / 10^{12})
7 = log(I_{1} / 10^{12}_{})
7 = log I_{1}  log 10^{12}_{}
log I_{1} = 7 + (12) I_{1} = 1X 10^{5} Wm^{2}
(b) β_{2} = (10 dB) log(I_{2} / I_{0})
50 dB = (10 dB) log(I_{2} / 10^{12}_{})
5 = log(I_{2} / 10^{12})
5 = log I_{2}  log 10^{12}_{}
log I_{2} = 5 + (12) I_{2} = 1X 10^{7} Wm^{2}
7. While standing near a railroad crossing, a person hears a distant train horn. According to the train's engineer, the frequency emitted by the horn is 440 Hz. The train is traveling at 20.0 ms^{1} and the speed of sound is 346 ms^{1}/s. (a) What would be the frequency of the train's horn if the train were at rest? (b) what is the apparent frequency that reaches the bystander as the train approaches the crossing? (c) What is the apparent frequency that reaches the bystander once the train has passed the crossing? Answer (a) 440 Hz. Answer (b)416. Hz. Answer (c) 467 Hz.
8. A plane electromagnetic wave traveling in the positive
direction of an x axis in vacuum has components E_{x} = E_{y} = 0 and E_{z} = (2.0 Vm^{1}) cos[(π X 10^{15} s^{1})(t  x/c)]. (a) What is the amplitude of the magnetic ﬁeld component? (b) Parallel to which axis does the
magnetic ﬁeld oscillate? (c) When the electric ﬁeld component is in
the positive direction of the z axis at a certain point P, what is the
direction of the magnetic ﬁeld component there? (a) The amplitude of the magnetic field is B_{m} = [E_{m} / c]
B_{m} = [(2.0 Vm^{1}) / (3.0 X 10^{8}ms^{1})] B_{m} = 6 .7 X 10^{9}T
(b) Since the Ewave oscillates in the z direction and travels in the x direction, we have B_{x} = B _{y} = 0. So, the oscillation of the magnetic field is parallel to the y axis.
(c) The direction (+x) of the electromagnetic wave propagation is determined by
EXB_{x}. If
the electric field points in +z, then the magnetic field must point in the –y direction.
Note:
1. The cross product EXB always gives the direction in which the wave travels (you know this as the right hand screw rule (also called righthand grip rule, coffeemug rule, or the corkscrewrule).
2. The electric and magnetic ﬁelds and are always perpendicular to the
direction in which the wave is traveling. Thus, the wave is a transverse wave.
3. The electric ﬁeld is always perpendicular to the magnetic ﬁeld.
4.
The ﬁelds always vary sinusoidally, with the same frequency and in phase
(in step) with each other.
9. A distant galaxy is moving away from us at approximately 5x10^{7}ms^{1} and we approximate the speed of light as c = 3x10^{8}ms^{1}. (a) What is the resulting wavelength of the Hydrogen spectral line of λ = 434 nm? (b) what would be the shift if the galaxy was moving towards earth? (a) f_{source} = ^{c}/_{λ}
The frequency has shifted from violet to ultraviolet.
10. Discuss the (a) photoelectric effect, (b) the Compton Effect and (c) Pair Production
1. The photoelectric effect  This happens when a photon knocks an electron out of an atom resulting in the disappearance of the photon.
— Directing a beam of light of certain short wavelength onto a clean metal surface causes electrons to be ejected from the surface.
— This photoelectric effect is used in many devices, including TV cameras, camcorders, and night vision viewers.
— This effect cannot be understood without quantum physics and Einstein used it to supported his photon concept.
— The energy of the photons is given by the equation E = hf, where h = Planck's constant and f the frequency of the radiation.
Experimental Evidence
— The maximum kinetic energy (of the most energetic ejected electrons) does not depend on the intensity of the light source. In classical physics the energy of the ejected electron should vary with the energy of the light wave (amplitude of the wave). Explanation: The energy of the photons is given by the equation E = hf (quantum mechanics), so the maximum energy given to an electron depends on the frequency and not the intensity (in classical physics)
—There is a cutoff frequency under which electrons are not ejected no matter how intense the radiation. In classical physics electrons should be ejected for all frequencies if the intensity is great enough. Explanation: The energy of the photons is given by the equation E = hf (quantum mechanics), so the maximum energy given to an electron depends on the frequency and not the intensity (in classical physics). To escape from the atom the electron must acquire a certain minimum energy called the work function (Φ). Electrons can only escape if hf > Φ.
Photoelectric Equation
— Einstein's statement on the conservation of energy for a single photon is the equation hf = E_{KMax} + Φ.
— To escape an e3lectron must pick up energy at least equal to Φ (hence the cutoff frequency) and the addition energy of the photon appears as E_{K} (hence there is a E_{KMax} which is the difference between the energy of the photon and Φ. Thus higher frequencies of the radiation produce more energetic photoelectrons.
— The slope of the graph on the left is [h/e]. From this graph (experimental results) the value of is the expected value as measured by other methods.
2. Excitation  The photon may knock an atomic electron to a higher energy state in the atom if its energy is not sufficient to knock the electron out altogether. In this process the photon also disappears, and all its energy is given to the atom. Such an atom is then said to be in an excited state, and we shall discuss it more later.
— Heated solids, liquids and dense gases emit light with a continuous spectrum of wavelengths. This radiation is assumed to be due to oscillations of atoms and molecules, which are largely governed by the interaction of each atom or molecule with its neighbors. Rarified gases on the other hand when excited to emit light do so at only certain wavelengths giving rise to line spectra rather than continuous spectra which are specific top the material. the opposite or absorption spectra are observed when continuous spectra are passed through the same materials.
— In rarified gases (low density) the light emitted or absorbed are by individual atoms rather than interacting atoms.
— The Balmer series of spectral lines for hydrogen is shown on the left. These fit the formula:
1
1
1
—
=
R
—

—
λ
2^{2}
n^{2}
where R is called the Rydberg constant. and n = 3.4, ....
The Lyman series (in the UV region) and Pashen series (in the IR region) show the same patterns.
— These can only be explained with quantum physics as the classical physics of the Rutherford model cannot explain them, quite apart from the obvious difficulties of a charged particle (electron) rotating in an electric field without emitting light and spiraling into the nucleus.
3. Compton Effect Sometimes the photon is scattered from an electron or nucleus lose some energy in the process. However, the photon is not slowed down as its speed is still c. As it has lost some energy its frequency must be lowered (recall E = hf).
XRay photon moving towards an electron.
XRay photon bypasses electron and no scatter (scattering angle = 0).
XRay photon scattered at angle φ with longer wavelength λ''and energy transferred to electron at angle θ.
XRay photon backscattered with longer wavelength λ' and maximum energy transferred to electron.
— Compton directed a beam of Xrays of a particular wavelength to a carbon target. The scattered Xrays contained a range of wavelengths with two prominent peaks for every angles observed.
— However, classical physics could not explain the scattering of xrays as the scattered rays should have the same wavelength and frequency as the incident rays.
— This has extended the concept of photons to also possess linear momentum. When a photon interacts with matter, it behaves as a collision in the classical sense with the conservation of Energy and Momentum.
4. Pair production: A photon can actually create matter, such as the production of an electron and a positron. (Recall: A positron has the same mass as an electron, but the opposite charge,)
— In pair production, the photon disappears in the process of creating the electron–positron pair. This is an example of
— Mass is being created from pure energy in accord with Einstein’s equation E = mc^{2} (the photon cannot create an electron only as electric charge would not then be conserved.)
— if a positron comes close to an electron, the two quickly annihilate each other and their energy, including their mass, appears as electromagnetic energy of photons (the inverse of pair production)
— Positrons are rarer than electrons nature so they do not last long.
— since the electron and positron move in the same direction along one of the axes, pair production must have a massive object like a nucleus to carry momentum in the opposite direction (Law of conservation of Momentum)
11. (a)Discuss the formation of images by plane and spherical mirrors.
(a)
/ Inverted
Image in Plane Mirrors.
Object distance
Image distance
Type of image
Size
Inverted
All
= Object distance
Virtual
Same
Lateral
/ Inverted
Image in Concave Mirrors.
Object distance
Image distance
Type of image
Size
Inverted
Object distance > c
f > Image distance < c
Real
Diminished
Yes
Object distance = c
Image distance = c
Real
Same
Yes
f > Object distance <c
c > Image distance
Real
Magnified
Yes
Object distance = f
∞
Real
Magnified
Yes
Object distance < f
Behind the mirror
Virtual
Magnified
No
/ Inverted
Image in Convex Mirrors.
Object distance
Image distance
Type of image
Size
Inverted
All
> f behind the mirror
Real
Diminished
No
12. Discuss the four major kinds of interactions when photons pass through matter and their applications in development of technology.
1. The photoelectric effect: This happens when a photon knocks an electron out of an atom resulting in the disappearance of the photon.
Directing a beam of light of certain short wavelength onto a clean metal surface causes electrons to be ejected from the surface.
— The energy of the photons is given by the equation E = hf, where h = Planck's constant and f the frequency of the radiation.
— The maximum kinetic energy (of the most energetic ejected electrons) does not depend on the intensity of the light source. In classical physics the energy of the ejected electron should vary with the energy of the light wave (amplitude of the wave). The energy of the photons is given by the equation E = hf (quantum mechanics), so the maximum energy given to an electron depends on the frequency and not the intensity (in classical physics)
—There is a cutoff frequency under which electrons are not ejected no matter how intense the radiation. In classical physics electrons should be ejected for all frequencies if the intensity is great enough. The energy of the photons is given by the equation E = hf (quantum mechanics), so the maximum energy given to an electron depends on the frequency and not the intensity (in classical physics). To escape from the atom the electron must acquire a certain minimum energy called the work function (Φ). Electrons can only escape if hf > Φ.
2. Excitation: The photon may knock an atomic electron to a higher energy state in the atom if
its energy is not sufficient to knock the electron out altogether. In this process
the photon also disappears, and all its energy is given to the atom. Such an
atom is then said to be in an excited state. — Heated solids, liquids and dense gases emit light with a continuous spectrum of wavelengths. This radiation is assumed to be due to oscillations of atoms and molecules, which are largely governed by the interaction of each atom or molecule with its neighbors. Rarified gases on the other hand when excited to emit light do so at only certain wavelengths giving rise to line spectra rather than continuous spectra which are specific top the material. the opposite or absorption spectra are observed when continuous spectra are passed through the same materials.
— In rarified gases (low density) the light emitted or absorbed are by individual atoms rather than interacting atoms.
— The Balmer series of spectral lines for hydrogen is shown on the left.
The Lyman series (in the UV region) and Pashen series (in the IR region) show the same patterns.
— These can only be explained with quantum physics as the classical physics of the Rutherford model cannot explain them, quite apart from the obvious difficulties of a charged particle (electron) rotating in an electric field without emitting light and spiraling into the nucleus.
3. Compton Effect: Sometimes the photon is scattered from an electron or nucleus lose some energy in the process. However, the photon is not slowed down as its speed is still c. As it has lost some energy its frequency must be lowered (recall E = hf).
— Compton directed a beam of Xrays of a particular wavelength to a carbon target. The scattered Xrays contained a range of wavelengths with two prominent peaks for every angles observed.
— However, classical physics could not explain the scattering of xrays as the scattered rays should have the same wavelength and frequency as the incident rays.
— This has extended the concept of photons to also possess linear momentum. When a photon interacts with matter, it behaves as a collision in the classical sense with the conservation of Energy and Momentum.
4. Pair production: A photon can actually create matter, such as the production of an electron and a positron. (Recall: A positron has the same mass as an electron, but the opposite charge,)
— In pair production, the photon disappears in the process of creating the electron–positron pair. This is an example of
— Mass is being created from pure energy in accord with Einstein’s equation E = mc^{2} (the photon cannot create an electron only as electric charge would not then be conserved.)
— If a positron comes close to an electron, the two quickly annihilate each other and their energy, including their mass, appears as electromagnetic energy of photons (the inverse of pair production)
— Positrons are rarer than electrons nature so they do not last long. — since the electron and positron move in the same direction along one of the axes, pair production must have a massive object (compared to the photon) like a nucleus to carry momentum in the opposite direction (Law of conservation of Momentum)
Some applications of the above effects are:
— Electron–positron annihilation is the basis for the type of medical imaging known as PET.
— Nuclear medicine
— Burglar alarms and automatic doors often make use of the photocell circuit
— Photocells are used in many devices, such as absorption spectrophotometers, to measure light intensity.
— Semiconductors