All questions of Sound (PHY) for MCAT Exam

F(s)/F(l) = [V+V(s)]/[V+V(l)]
V = velocity of sound = 340m/s
V(l) = velocity of listener = 0
F(l) = frequency heard by listener = ?
V(s) = velocity of source = -20m/s (because, it's source to listener)
F(s) = frequency of source = 600hz
By putting these values in the above formula and solving we get,
F(l) = 637.5 Hz.
Hence C is correct.

We need to find the resulting wavelength of the sound wave, which can be calculated using the formula: λ' = λ(v +/- vs) / (v + u) Since the source is moving towards the observer, we can use the negative sign for vs: λ' = λ(v - vs) / (v + u) λ' = λ(v - (-1/3)v) / (v + u) λ' = λ(4/3v) / (v + u) We can also use the formula for the speed of the sound wave: v = fλ which gives us: λ = v/f Substituting this value in the above equation, we get: λ' = (4/3)(v/f) / (v + u) λ' = (4/3)f / (3v + u) We know that the velocity of the observer (u) is zero since it is stationary. Thus, the equation simplifies to: λ' = (4/9)f/v This means that the resulting wavelength is 2/3 of the emitted wavelength since: λ' / λ = (4/9f/v) / (f/v) = 4/9 = 0.44 Therefore, the correct answer is option B, 2/3 of the emitted wavelength.

The reason for the Doppler effect is that when the source of the waves is moving towards the observer, each successive wave crest is emitted from a position closer to the observer than the crest of the previous wave.

Given:
Frequency of radio signal n=9×10 ⁹Hz
 Frequency shift n0=3×10³Hz
 Velocity of radio signal v=3×10⁸m/s
 Frequency shift shown by reflected wave is Shift =n−n
=>n−n=(v+u/v−u)n−n
 
Shift=(2u/v−u)ⁿ
Now, putting the given values in Eq. (i), we get,
3×10³=(2u/3×10^-8−u)9×10⁹
                    
⇒3×10⁸−u=2u×9×10⁹/3×10³
=6u×10⁶
⇒6×10⁶u+u=3×10⁸
 
u(6×106+1)=3×10⁸
 
u(6000001)=3×10⁸
 
u=3×10⁸/6000001=49.999≈50m/s

The Doppler Effect is observed whenever the speed of the source is moving slower than the speed of waves. But if the source actually moves at the same speed as or faster than the waves, a different phenomenon is observed. This phenomenon is known as Shock waves or Sonic Booms.

When no beats is listened at the station thus we get the relative frequencies of both the trains are the same.
Hence the relative frequency of the train B is 600 (330/300) = 660
similarly the relative frequency of train A is f(330/315) = 660
Hence we get f = 630Hz

The angular frequency is related to the frequency through the equation ω = 2πf. Therefore, the magnitude of the angular frequency will always be larger than the magnitude of the frequency. The magnitude of the angular frequency may or may not be larger than the magnitude of period; these variables are inversely proportional, eliminating choice (A). The product of the frequency and the period is always 1 because these two are inverses of each other, eliminating choice (B). Finally, the product of the angular frequency and period will always be 2π because ω = 2πf = 2π/T, eliminating choice (D).

This question is testing us on our understanding of the Doppler effect. A difference of zero between the perceived and the emitted frequencies implies that the source of the sound is not moving relative to the student. If the car in choice (D) is moving at the same speed as the student, then the relative motion between them could be 0. In all of the other cases, the student and the sound source are necessarily moving relative to each other.

Saying that the stapedial footplate has limited displacement during vibration is another way of stating that the amplitude of the vibration has been decreased. Because amplitude is related to intensity, and intensity is related to sound level, the perceived sound level (volume) will be decreased as well. Pitch, described in choices (C) and (D), is related to the frequency of a sound, not its amplitude.

Intensity is equal to power divided by area. In this case, area refers to the surface area of concentric spheres emanating out from the source of the sound. This surface area is given by 4πr², so as distance (r) doubles, the intensity will decrease by a factor of four.

Period is inversely related to frequency. Because the perceived frequency is doubled, the perceived period must be halved, from 34 ms to 17 ms. While either condition I or II would cause a doubling of the perceived velocity, neither condition must necessarily be true because the opposite could be true instead.

This question is testing our understanding of pipes open at one or both ends. To begin, remember that high-frequency sounds have a high pitch and low-frequency sounds have a low pitch. The pipe in this example begins as one that is open on both ends, and then one end is closed off. Our task, therefore, is to determine how the frequency of the second harmonic differs between a pipe that is open at both ends from one of equal length that is open at only one end. For a pipe of length L open at both ends, the wavelength for the second harmonic (first overtone) is equal to L:

In contrast, for a pipe open at one end and closed at the other, the wavelength is equal to 4L/3:

Keep in mind that the first overtone for a closed pipe corresponds to the third harmonic, not the second. Thus, when the brother covers one end of the flute, the wavelength increases. Given that the wavelength and the frequency of a sound are inversely proportional, an increase in wavelength corresponds to a decrease in frequency. Therefore, when the brother covers one end of the flute, the sound produced by the instrument will be slightly lower in pitch than the original sound.

While intensity, choice (A), could be used to measure distance, time of travel is an easier indication and most commonly used by ultrasound machines. Apparent frequency, choice (D), is only used in Doppler ultrasound, and not to calculate distance. Angle of incidence, choice (C), can be used to position various structures on the screen of an ultrasound, but is not used to calculate distance.

The angular frequency is related to the frequency of a wave through the formula ω = 2πf. Thus, our initial task is to calculate the frequency of the wave. Knowing its speed, we determine the frequency by first calculating its wavelength (ν = fλ). For the third harmonic of a standing wave in a pipe with one closed end, the wavelength is

The frequency of the wave is therefore

Finally, obtain the angular frequency simply by multiplying the frequency of the wave by 2π:
ω = 2πf = 850π radians per second

If these two glasses are perfectly identical, then the fact that the first glass shattered at 808 Hz tells us that this is very close (if not identical) to the natural (resonant) frequency of the glass. If she produces a frequency that is not equal (or very close) to the natural frequency, then the applied frequency will not cause the glass to resonate, and there will not be the increase in wave amplitude associated with resonating objects. Attenuation will increase with increased frequency because there is more motion over which nonconservative forces can damp the sound wave; however, even if sound level was matched to that which shattered the first glass when accounting for attenuation, the glass would still not shatter for the reasons described above, eliminating choice (D).

Shock waves are the buildup of wave fronts as the distance between those wave fronts decreases. This occurs maximally when an object is traveling at exactly the same speed as the wave is traveling (the speed of sound). Once an object moves faster than the speed of sound, some of the effects of the shock wave are mitigated because all of the wave fronts will trail behind the object, destructively interfering with each other.