Unit 1 Worksheet
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Unit 1.3.5 Doppler Effect Worksheet
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Unit 2 Worksheet
Unit 2 Worksheet Answers
Final Worksheet
Alternate Worksheet
Worksheet 2016
Worksheet 2016 Answers

Worksheet 2016

1. (a) Discuss how loudness is measured.
(b) The sound in the school yard is 20 dB. At Carnival time a music truck is measured at 50dB in the same school yard. How many more times is the intensity of the sound from the music truck?

2. 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) β1 = (10 dB) log(I1 / I0)
70 dB = (10 dB) log(I1 / 10-12)
7 = log(I1 / 10-12)
7 = log I1 - log 10-12
log I1 = 7 + (-12)
I1 = 1X 10-5 Wm2

(b) β2 = (10 dB) log(I2 / I0)
50 dB = (10 dB) log(I2 / 10-12)
5 = log(I2 / 10-12)
5 = log I2 - log 10-12
log I2 = 5 + (-12)
I2 = 1X 10-7 Wm2


q93. 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 10-3kgm-1)]
v = 1.44 X 102 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 102 ms-1) / (0.60 m)
f = 240 Hz.

4. (a) Find the speed of waves on a violin string of mass 800 mg and length 22.0 cm if the fundamental frequency is 920 Hz. (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string?

(a) When the string (fixed at both ends) is vibrating at its lowest resonant frequency, exactly one-half of a wavelength fits between the ends. 
Thus, λ = 2L. 
Now v = fλ
v = 2Lf 
v = 2(0.220 m)(920 Hz) 
v = = 405 ms-1.
(b) 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. 
If M is the mass of the (uniform) string, then μ = M/L. 
Thus, τ = μv2
τ = (M/L)v2
τ = [(800 × 106 kg)/(0.220 m)] (405 ms-1)2
τ = 596 N.
(c) The wavelength is λ = 2L 
λ = 2(0.220 m) 
λ = 0.440 m.
(d) The frequency of the sound wave in air is the same as the frequency of oscillation of the string. The wavelength is different because the wave speed is different. If v is the speed of sound in air, the wavelength in air is 
λa = [va / f] 
λa = (343 m/s)/(920 Hz) 
λa = 0.373 m.

5. Two identical cars are driving toward one another and sounding their horns. You're the driver of one of the cars. You measure your car's horn to be sounding at 512 Hz, but you measure the horn of the other car to be sounding at 600. Hz. The speed of sound is 345 ms-1. If you are traveling at 26.8 ms-1, how fast is the other car traveling?

fo = fs [(v ± vo) / (v ± vs)]

600 Hz = 5.12 X 102 Hz [(345 ms-1+ 26.8 ms-1) / (345 ms-1 - vs)]

(343 ms-1 - vs) (600 Hz) = 5.12 X 102 Hz [(345 ms-1+ 26.8 ms-1)

(343 ms-1 - vs) = 5.12 X 102 Hz [(345 ms-1+ 26.8 ms-1) / (600 Hz)

vs= 343 ms-1 - [5.12 X 102 Hz (345 ms-1+ 26.8 ms-1) / (600 Hz)]

vs = 25.7 ms-1

6. A stationary motion detector sends sound waves of frequency 0.150 MHz and speed 343 ms-1 toward a truck approaching at a speed of 45.0 ms-1. What is the frequency of the waves (a) detected by the truck, (b) reflected back to the detector?

(a) fo = fs [(v ± vo) / (v ± vs)]

For a stationary source and a moving observer towards the source

fo = fs [(v + vo) / (v - vs)]
fo = 1.50 X 105 Hz [(343 ms-1+ 45.0 ms-1) / (343 ms-1 - 0)]
fo = 1.70 X 105 Hz

(b) The reflected wave from the truck now as a moving wave to be reflected back to the stationary detector.
For a moving source towards the observer and a stationary observer fo = fs [(v + vo) / (v - vs)]
fo = 1.70 X 105 [(343 ms-1+ 0) / (343 ms-1 - 45.0 ms-1)]
fo = 1.95 X 105 Hz

7. Discuss the (a) photoelectric effect, (b) the Compton Effect and (c) Pair Production.

1. The photoelectric effect -

— 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 cut-off 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 = EKMax + Φ.
— To escape an e3lectron must pick up energy at least equal to Φ (hence the cut-off frequency) and the addition energy of the photon appears as EK (hence there is a EKMax which is the difference between the energy of the photon and Φ. Thus higher frequencies of the radiation produce more energetic photo-electrons.
— 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      22   n2

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).
X-Ray photon moving towards an electron.
X-Ray photon bypasses electron and no scatter (scattering angle = 0).
X-Ray photon scattered at angle φ with longer wavelength λ''and energy transferred to electron at angle θ.
X-Ray photon backscattered with longer wavelength λ' and maximum energy transferred to electron.

— Compton directed a beam of X-rays of a particular wavelength to a carbon target. The scattered X-rays contained a range of wavelengths with two prominent peaks for every angles observed.

— However, classical physics could not explain the scattering of x-rays 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 = mc2 (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)



Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.