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All Solutions of Type: Wave Phenomena
 0 Click here to jump to the problem! GR8677 #4 Wave Phenomena$\Rightarrow$}Wave Equation Perhaps this formula is more familiar: $y=A\sin(\omega t - k x)$. But then again, they define the familiar quantities $k$ and $\omega$ for you in theirs... (A) The amplitude is A. (Recall that the amplitude of $y=sin(x)$ is just 1, not 2.) (B) How exactly does the argument of $\sin$ make it a traveling wave? Well, a traveling wave keeps the same waveform at all times. Thus, the overall argument has to be constant. $\omega t - kx = constant\Rightarrow x \propto - constant +\omega t$. This looks like the old high school kinematics equation $x=-constant + vt$. In fact, it's basically the same thing. That high school equation describes a particle going in the positive x-direction. So does this equation. (C) Dimensions don't match. Period has units of time, not units of time/meter. (D) The speed of the wave is $v=\lambda/T$. From the kinematics explanation explained in (B), the velocity of the wave is obviously $v=\frac{\omega}{k}=\frac{2\pi\lambda}{2\pi T}$ (E) True. See (D) and (B). Click here to jump to the problem!
 1 Click here to jump to the problem! GR8677 #12 Wave Phenomena$\Rightarrow$}Doppler Effect For a source that's coming towards you, its frequency will obviously be higher than its original source frequency. Choices A and B are out. To know for sure whether the choice is C, D, or E, one can apply the Doppler Effect. The Doppler Effect equation can easily be derived even if one forgets it. A source $f_0$ approaching at $v_0$ would have $\Delta \lambda$, the distance between waves, decreasing. $v=\lambda f\Rightarrow \Delta \lambda=\frac{v_{sound}-v_{0}}{f_{0}}$. The $\Delta \lambda$ wave travels at the speed of sound, thus the frequency you receive is: $f=\frac{v_{sound}}{\Delta \lambda}=10f_0$. Choice E. Click here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #59 Wave Phenomena$\Rightarrow$}Group Velocity Group velocity is $v_g=d\omega/dk$. So, take the derivative of the quantity $\omega = \sqrt{c^2 k^2 + m^2}$ to get $v_g=d\omega/dk = \frac{c^2k}{\sqrt{c^2 k^2 + m^2}}. $ Use the above equation to test the 5 choices: (A) As $k \rightarrow 0 \Rightarrow \omega = 0$, not infinity. The first condition doesn't work, no need to test the second (don't have to remember L'Hopital's rule). (B) Wrong for the same reason as (A). (C) Wrong because $v_g$ approaches $0$, not $c$, as $k \rightarrow 0$. (D) As $k \rightarrow \infty$, $v_g \approx \frac{kc^2}{\sqrt{c^2 k^2}}=c$, since $m \ll (ck)^2$. So, $v_g$ doesn't tend towards $\infty$. This choice is wrong. (E) This is it. The conditions work (and it's the only choice left). Click here to jump to the problem!
 3 Click here to jump to the problem! GR8677 #72 Wave Phenomena$\Rightarrow$}Group Velocity Recall the definition of index of refraction $n$ in terms of the speed of light in vacuum $c$ and the velocity of the light in medium $v$, $n=c/v$. Since the velocity of light in any medium is $v\leq c$, the condition $n \geq 1$ usually holds. However, even if rock salt has $n > 1$, the wave does not exceed the speed of light. The group velocity can travel faster than the speed of light, and apparently it is the group velocity at work in the equation $n=c/v$. One can also arrive at this conclusion via MOE (Method of Elimination): (A) Relativity is a pretty general theory that supposedly applies everywhere. (Newtonian mechanics can be achieved using the proper approximation technique.) X-out this choice. (B) An x-ray is a specific frequency in the electromagnetic spectrum. Nothing forbids an electromagnetic wave from transmitting signals, and thus this choice is out. (C) Imaginary mass? If confused, save for last comparison. (D) Historical precedence shouldn't change the correctness of a theory (at least not in the ideal world ETS lives in)... X-this out. (E) One recalls that there's a difference between group and phase velocities. Could this difference allow one of them to exceed the speed of light? Probably. In either case, this is a much better choice than the other remaining candidate, choice (C). So, choose this. Click here to jump to the problem!
 4 Click here to jump to the problem! GR8677 #77 Wave Phenomena$\Rightarrow$}Wave Equation The problem gives the equation of motion \par $m\ddot{x}=-kx \Rightarrow \ddot{x}=-\frac{k}{m}x=-\omega^2 x$, where $\omega^2=\frac{k}{m}$. The general equation for a wave propagating in time and oscillating in the x direction is $x(t)=A \sin(\omega t + \phi)$. ($A$ is the amplitude, $\omega$ is the angular frequency and $\phi$ is some phase constant.) This is also the general solution to the differential equation posed above. Plug in the condition (given by the problem) that \par $x(t)=A/2=A\sin(\omega t +\phi)$ to get $1/2=\sin(\omega t + \phi)$. Recalling the unit circle, the angle \par $\omega t + \phi = \pi/6$. Plug in the argument into the velocity \par $\dot{x}=\omega A \cos(\omega t+\phi)=\omega A \cos(\pi/6)=\omega A \sqrt{3}/2$. Recall that $\omega = 2\pi f$, and thus $\dot{x}=\pi f A\sqrt{3}$, as in choice (B). Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #60 Wave Phenomena$\Rightarrow$}Light Doppler Shift One can derive the Doppler Shift for light as follows: For source/observer moving towards each other, one has the wavelength emitted from the source decreasing, thus $\lambda = (cdt - vdt) = (c - v) t_0\gamma$. Thus, $\lambda = (c-v)\gamma\lambda_0/c$. For source/observer moving away from each other, one has the wavelength emitted from the source increasing, thus $\lambda = (cdt + vdt)=(c + v) t_0\gamma$. Thus, $\lambda = (c+v)\gamma\lambda_0/c$. Where in the last equality in the above, one applies time dilation from special relativity, $t = t_0 \gamma$ and the fact that $c=\lambda f = \lambda/t$ in general. Now that one has the proper battle equipment, one can proceed with the problem. This problem is essentially the difference in wavelengths seen from a red shift and blue shift, i.e., light moving towards and away from the observer. $\Delta \lambda = \left( 2v \right)\gamma\lambda_0/c \Rightarrow v = \frac{\Delta \lambda c}{2 \gamma \lambda_0} \approx \frac{\Delta \lambda c}{2 \lambda_0} \approx \frac{1.8E-12 \times 3E8}{244 E-9}\approx 2E3$, where the approximation $\gamma \approx 1$ is made since one assumes the particle is moving at a non-relativistic speed. 2 km is closest to choice (B). Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #81 Wave Phenomena$\Rightarrow$}Beats One remembers the relation for beats, i.e., $f_1-f_2=f_{beat}$. Beat phenomenon occurs when two waves occur at nearly the same frequency. Taking $f_0>\approx 73$ as the fundamental frequency, one deduces the harmonic to be $440/f_0\approx 6$. $440-6f_0\approx 0.5$, and thus the answer is choice (B). One can derive the relation for beats by recalling the fact that one gets beat phenomenon when one superposes two sound waves of similar frequency $f_1\approx f_2$, say, of the form $A \sin(2 \pi f_i t)$, $f=A \sin(2 \pi f_1 t)+A \sin(2 \pi f_2 t)=2A\sin(2\pi(f_1+f_2)/2 t)\cos(2\pi(f_1-f_2)/2 t), $ where to get beats phenomenon one must have $\cos(t (f_1-f_2)/2)=\pm 1 \Rightarrow 2\pi = (f_1-f_2)/2 t$, and since there are two beats per period, one has $f_1-f_2=f_{beat}$. Click here to jump to the problem!
 7 Click here to jump to the problem! GR8677 #50 Wave Phenomena$\Rightarrow$}Sound Since the wavelength of the wave does not change, as the pipe presumably stays approximately the same length, only the frequency varies. If the speed of sound changes, then the frequency changes. If the speed of sound is lower than usual, then the frequency is lower. Thus, choices (A), (B) and (C) remain. Calculate $0.97 \times 440$ to get choice (B). Click here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #79 Wave Phenomena$\Rightarrow$}Group Velocity Recall that the group velocity is given by $v_g = \frac{d\omega}{dk}$ and the phase velocity is given by $v_p = \omega/k$. In the region between $k_1$ and $k_2$, the derivative is a constant negative quantity (approximately just the derivative of a line with negative slope). However, $\omega/k$ is positive in this region. Thus, the phase and group velocity are traveling in opposite directions. Thus, choose choice (A). Click here to jump to the problem!

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