sound khan academy
Compare the frequencies of the third harmonics between a closed-pipe (f_c and open-pipe (f_o instrument. Which is true? Choose 1 answer: f_c> f_o by a factor of 4 f_c< f_o by a factor of 4 f_c> f_o by a factor of 2 f_c< f_o by a factor of 2
Recall that for a closed-pipe, L(the first harmonic wavelength) = ¼ of the wavelength, and that for an open pipe, L = ½ of the wavelength. Hint #22 / 3 Use speed of sound = frequency*wavelength. Hint #33 / 3 We can see that fundamental frequency for closed pipe = speed of sound / 4L, and that for an open pipe = speed of sound / 2L. The third harmonic would be just the fundamental frequency*3. Therefore, we can see that the frequency for a closed pipe is half of that for an open pipe. Therefore, fc < fo, by a factor of 2 is correct.
Which statement below is true? Choose 1 answer: Choose 1 answer: Sound waves can propagate as longitudinal or transverse waves, depending on the temperature. Sound waves are transverse waves and they propagate perpendicular to the transmitting medium. Sound waves can propagate as longitudinal or transverse waves, depending on the transmitting medium. Sound waves are longitudinal waves and they propagate parallel to the transmitting medium.
Sound waves are longitudinal waves and they propagate parallel to the transmitting medium. Longitudinal waves propagate parallel; transverse waves propagate perpendicularly. Hint #22 / 3 Transverse waves can only occur on a string, on the surface of a liquid, or through a solid. Hint #33 / 3 Sound waves are longitudinal waves and the propagate parallel to the transmitting medium. Although this seems like a simple definition question, it is important to realize that a sound wave must be a longitudinal wave because there is no way for a transverse wave to propagate inside a gas or a liquid in a way to maintain the perpendicular driving motion.
A car travelled traveling at 10 m/s towards you continuously plays music that has an average frequency of 150 Hz (this is roughly the note D). What frequency (in Hz) do you hear as she approaches? What frequency do you hear as she passes by? (speed of sound in air in m/s = s) Choose 1 answer: Choose 1 answer: Approaching: 150 \times×times (\dfrac{s}{s+10})( s+10 s )left parenthesis, start fraction, s, divided by, s, plus, 10, end fraction, right parenthesis; leaving: 150 \times×times (\dfrac{s}{s-10})( s−10 s )left parenthesis, start fraction, s, divided by, s, minus, 10, end fraction, right parenthesis Approaching: 150 \times×times (\dfrac{s-10}{s})( s s−10 )left parenthesis, start fraction, s, minus, 10, divided by, s, end fraction, right parenthesis; leaving: 150 \times×times (\dfrac{s+10}{s})( s s+10 )left parenthesis, start fraction, s, plus, 10, divided by, s, end fraction, right parenthesis Approaching: 150 \times×times (\dfrac{s+10}{s})( s s+10 )left parenthesis, start fraction, s, plus, 10, divided by, s, end fraction, right parenthesis; leaving: 150 \times×times (\dfrac{s-10}{s})( s s−10 )left parenthesis, start fraction, s, minus, 10, divided by, s, end fraction, right parenthesis Approaching: 150 \times×times (\dfrac{s}{s-10})( s−10 s )left parenthesis, start fraction, s, divided by, s, minus, 10, end fraction, right parenthesis; leaving: 150 \times×times (\dfrac{s}{s+10})( s+10 s )left parenthesis, start fraction, s, divided by, s, plus, 10, end fraction, right parenthesis
When the car stereo passes you, you will first hear a higher pitched sound, then a lower pitched sound when it passes. Hint #22 / 3 Quantitatively, the Doppler frequency when a car passes can be found using: f_D D start subscript, D, end subscript = f \times×times \dfrac{s}{(s-speed_{car})} (s−speed car ) s start fraction, s, divided by, left parenthesis, s, minus, s, p, e, e, d, start subscript, c, a, r, end subscript, right parenthesis, end fraction during the approach, and f_D D start subscript, D, end subscript = f \times×times \dfrac{s}{(s+speed_{car})} (s+speed car ) s start fraction, s, divided by, left parenthesis, s, plus, s, p, e, e, d, start subscript, c, a, r, end subscript, right parenthesis, end fraction during the exit. This is consistent with higher frequency during the approach and lower frequency during the exit.
The speed of sound traveling through a gaseous sample of mixed gases can be affected by: I. Temperature II. Density III. Pressure IV. Composition of the gaseous sample Choose 1 answer: Choose 1 answer: One of the above Two of the above Three of the above All of the above
all of the above. Imagine the propagation of sound through a medium as passing energy through an array of balls connected by numerous springs. This is an imperfect model, but consider the factors that allow the energy to be propagated more quickly. Hint #22 / 3 Composition of the gaseous sample affects the density of the gas. Also, speed of sound through a medium = square root(elastic modulus/density). Hint #33 / 3 The speed of sound traveling through a gaseous sample of mixed gases is affected by temperature, density, pressure, and composition of the gaseous sample. For example, the speed of sound is faster at higher temperatures (imagine tighter springs allowing the balls in our model to vibrate faster, and thus allowing sound to propagate through the medium quicker) and higher pressure (roughly the same reasoning). The answer is all of the above.
A sound wave of wavelength λ and frequency f is propagated through a gold rod, then exits into the surrounding air. What most accurately describes the relationship between the original wavelength and frequency with the new wavelength λ' and new frequency f' in air? Choose 1 answer: Choose 1 answer: λ > λ', f = f' λ = λ', f < f' λ < λ', f = f' λ = λ', f > f'
λ > λ', f = f' Sound travels faster in a solid than in a gas due to differences in the conducting medium. Hint #22 / 3 Wavelength is affected by changes in the conducting medium. Hint #33 / 3 Since frequency is only dependent on the initial sound source, it does not change with change in medium. Since sound travels faster in the solid, it will have a higher wavelength to account for the change in the speed (since frequency is unchanged). Therefore, λ > λ', f = f'.
