Electromagnetic Waves AP Physics 2 FRQ – Exam Style Questions etc.
Electromagnetic Waves AP Physics 2 FRQ
Unit 14: Waves , Sound , and Physical Optics
Weightage : 15–18%
Exam Style Practice Questions , Electromagnetic Waves AP Physics 2 FRQ
Question
The figure above represents the electric field \(E\) as a function of position \(x\) due to a particular radio wave at time \(t=0 \mathrm{~s}\). Which of the following equations correctly describes the electric field?
(A) \(E=(100 \mathrm{~N} / \mathrm{C}) \cos \left(\frac{\pi x}{1 \mathrm{~m}}\right)\)
(B) \(E=(100 \mathrm{~N} / \mathrm{C}) \cos \left(\frac{\pi x}{2 \mathrm{~m}}\right)\)
(C) \(E=(100 \mathrm{~N} / \mathrm{C}) \cos \left(\frac{\pi x}{4 \mathrm{~m}}\right)\)
(D) \(E=(100 \mathrm{~N} / \mathrm{C}) \sin \left(\frac{\pi x}{4 \mathrm{~m}}\right)\)
▶️Answer/Explanation
Ans: C
Questions
The figure above shows a model of an electromagnetic wave, where E is the electric field and B is the magnetic field. In what direction is the energy of the wave transmitted?
(A) Along the x-axis only
(B) Along the y-axis only
(C) Along the z-axis only
(D) In a direction that is at a nonzero angle to each of the axes
▶️Answer/Explanation
Ans: A
This option is correct. Electromagnetic waves, like waves on a string, are transverse: the direction of the “disturbance” is perpendicular to the direction in which the waves travel, i.e., the direction in which energy is transmitted. Here, the electric and magnetic “disturbances” are y-directed and z-directed; therefore the waves must be traveling perpendicular to both of those transverse disturbances, along the x-axis.
Question
You and a friend are traveling in a car. You tune the car radio to a station of frequency 850 kHz. The graph below represents the electric field strength of the radio wave at a given position as a function of time.
Which of the following best represents the electric field strength E measured in V/m as a function of time t measured in \(\mu s\)?
(A) \(E=(3.0\times 10^{-4})\sin (10.6t)\)
(B) \(E=(3.0\times 10^{-4})\sin (5.32t)\)
(C) \(E=(1.5\times 10^{-4})\sin (10.6t)\)
(D) \(E=(1.5\times 10^{-4})\sin (5.32t)\)
Answer/Explanation
Ans:D
Question
An observer can hear sound from around a comer but cannot see light from around the same comer. Which of the following helps to explain this phenomenon?
(A) Sound is a longitudinal wave, and light is an electromagnetic wave.
(B) Sound is a mechanical wave, and light is a transverse wave.
(C) Light travels at a speed much faster than that of sound.
(D) Light has a much smaller wavelength than sound.
Answer/Explanation
Ans:D
The point source model shows us that the larger the wavelength, the greater the bending of the wave around the corner. We can also see this in the equation for diffraction: \(\sin\theta=\frac{m\lambda}{d}\). As the wavelength gets smaller, the diffraction bending angle also gets smaller. Visible light has a very small wavelength and only bends a tiny amount around corners.
Question
What is the equation for the electric field given by the above plot?
The electric field crosses the axis at times o ps, 1.43 ps, 2.85 ps, and 4.28 ps.
(A) \(E = \left ( \frac{0 N}{C} \right )sin ((2.85 THz)t)\)
(B) \(E = \left ( \frac{10 N}{C} \right )sin ((2.85 THz)t)\)
(C) \(E = \left ( \frac{10 N}{C} \right )sin ((2.20 THz)t)\)
(D) \(E = \left ( \frac{10 N}{C} \right )cos ((0.45 THz)t)\)
Answer/Explanation
Ans:C
The field is a sine curve with an amplitude of 10 N/C. From the graph, the period of the wave is 2.85ps. The argument of sine needs to be 2πt/T.