- A Wave Rule #2 states that when a wave travels from one medium into another
medium the frequency remains the same. This eliminates (B), (D), and (E).
When the wave enters the air from the metal rod, its speed will decrease.
The frequency, however, will not change. Since v = λf must always be
satisfied, a decrease in v implies a decrease in λ.
- A The distance from S 2 to P is 5 m (it’s the hypotenuse of a 3-4-5 triangle),
and the distance from S 1 to P is 4 m. The difference between the path lengths
to point P is 1 m, which is half the wavelength. Therefore, the sound waves
are always exactly out of phase when they reach point P from the two
speakers, causing destructive interference there. By contrast, since point Q is
equidistant from the two speakers, the sound waves will always arrive in
phase at Q, interfering constructively. Since there’s destructive interference
at P and constructive interference at Q, the amplitude at P will be less than at
Q.
- E The intensity (power per unit area) is proportional to 1/r^2 , where r is the
distance between the source and the detector. If r increases by a factor of 10,
the intensity decreases by a factor of 100. Because the decibel scale is
logarithmic, if the intensity decreases by a factor of 100 = 10^2 , the decibel
level decreases by 10 + 10 = 20 dB.
- B An air column (such as an organ pipe) with one closed end resonates at
frequencies given by the equation fn = for odd integers n. The
fundamental frequency corresponds, by definition, to n = 1. Therefore
- B The speed of the chirp is
v = λf = (8.75 × 10−3 m)(40 × 10^3 Hz) = 350 m/s
If the distance from the bat to the tree is d, then the wave travels a
total distance of d + d = 2d (round-trip distance). If T is the time for
this round-trip, then
