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Sound Test - 35

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Sound Test - 35
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  • Question 1
    1 / -0
    Compression is the region of
    Solution

    Compression is the region of high pressure and high density.
    Region of high pressure and high density is called compression and region of low pressure and low density in the air is called rarefaction. 

  • Question 2
    1 / -0
    A sound wave has a frequency of 192 Hz and travels the length of a football field, 91.4 m, in 0.27 s. What is the wavelength of the wave?
    Solution
    $$Speed\ of\ sound=\dfrac { Distance\  traveled\ in\ m }{ Time\  taken\ in\ s } $$.
    In the question, it is given that the distance travelled$$(s)$$ is 91.4 m (length of the football field) and the time taken$$(t)$$ is 0.27 seconds.

    The speed of the sound wave is calculated as follows.
    $$v=\dfrac { s }{ t } \quad =\quad \dfrac { 91.4\ m }{ 0.27\ s } \quad =\quad 338.5\ m/s$$

    Hence, the speed of the sound wave is 338.5 m/s.
    Now the wavelength is calculated from the below relation .
    The relationship between the velocity of the sound, its wavelength and frequency is given by the formula $$Wavelength\quad \lambda =\dfrac { Speed\quad v }{ Frequency\quad \nu  } $$,
    where,
    $$v$$ - velocity of the wave
    $$\lambda $$- wavelength of the wave
    f - frequency of the wave.
    In the question it is given that the frequency$$(\nu)$$ is 192 Hz and the speed$$(v)$$ is 338.5 m/s(from the above calculation).
    That is, $$\lambda =\dfrac { v }{ \nu  } =\dfrac { 338.5\quad m/s }{ 192\quad Hz } =1.76\quad m$$

    Hence, the wavelength of the sound wave is 1.76 meters.
  • Question 3
    1 / -0
    A sound wave has a frequency of 192 Hz and travels the length of a football field, 91.4 m, in 0.27 s. What is the period of the wave?
    Solution
    The frequency (f) of a wave is the number of full wave forms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.  
    Time Period (T) is the number of seconds per waveform, or the number of seconds per oscillation.  It is clear that frequency and period are reciprocals.
    That is, $$T=\dfrac { 1 }{ f } =\dfrac { 1 }{ 192 } =0.0052\quad seconds$$.
    Hence, the time period of the wave is $$5.2\times { 10 }^{ -3 }\quad seconds$$.
  • Question 4
    1 / -0
    Echocardiography is used to diagnose heart diseases which utilises infrasonic waves.
    Solution
    An echocardiogram (also called an echo) is a type of ultrasound test that uses high-pitched sound waves that are sent through a device called a transducer. The device picks up echoes of the sound waves as they bounce off from the different parts of the heart. These echoes are turned into moving pictures of the heart that can be seen on a video screen.
    Hence, the statement is false.
  • Question 5
    1 / -0
    A hiker shouts toward a vertical cliff 465 m away. The echo is heard 2.75 s later. The wavelength of the sound is 0.750 m. What is its frequency?
    Solution
    Given,
    Distance of cliff $$d=465\ m/s$$
    Time taken to hear echo $$t=2.75\ sec$$
    Wavelength of sound $$\lambda=0.75\ m$$
    Total distance travel by the sound $$s=2\times d=2\times 465=930$$
    The speed of the sound, distance travelled and time is related with the following equation.
    $$Speed\; of\; sound=\dfrac {Distance\; traveled }{ Time\; taken }$$.

    The speed of the sound wave is calculated as follows.
                    $$v=\dfrac { s }{ t } = \dfrac { 930 m }{ 2.75 s } = 338.2m/s$$.
    The speed of the sound is 338.2 m/s.

    The relationship between frequency, speed of sound and wavelength of a sound wave is given as $$Frequency=\dfrac { Speed\; of\; sound\ (v)\; m/s }{ Wavelength\ (\lambda)\; m } $$.

    Therefore, $$Frequency\; \nu =\dfrac { 338.2\; m/s }{ 0.750\; m } \; =450Hz$$,

    Hence, the frequency of the sound wave is 450.9 Hz.

  • Question 6
    1 / -0
    In SONAR, we use
    Solution
    Answer is A.

    Sonar illustrates how a ship on the ocean utilizes the reflecting properties of ultrasound waves to determine the depth of the ocean. A ultrasound wave is transmitted and bounces off the seabed. Because the speed of sound is known and the time lapse between sending and receiving the sound can be measured, the distance from the ship to the bottom of the ocean can be determined. This technique is called sonar (originally an acronym for Sound Navigation And Ranging).
  • Question 7
    1 / -0
    Water waves in a shallow dish are $$6 cm$$ long. At one point, the water moves up and down at a rate of $$4.8$$ oscillations/s. What is the time period of the water waves?
    Solution
    The frequency (f) of a wave is the number of full wave forms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.  
    Time Period (T) is the number of seconds per waveform, or the number of seconds per oscillation.  It is clear that frequency and period are reciprocals. 
    That is, $$T=\dfrac{1}{f}$$

    $$\dfrac{1}{f}=\dfrac{1}{4.8}=0.208\;seconds$$
    Hence, the time period of the water wave is 0.208 seconds.
  • Question 8
    1 / -0
    In case of longitudinal waves, the particles of medium vibrate
    Solution
    Longitudinal waves, also known as l-waves, are waves in which the displacement of the medium is in the same direction as the direction of travel of the wave. Longitudinal waves are also called compressional waves or compression waves, because they produce compression and rarefaction when traveling through a medium.
  • Question 9
    1 / -0
    Which characteristic of a sound helps you to identify your friend by his voice while sitting with others in a dark room?
    Solution
    The sound waves given off by different vibrating bodies differ in quality or timbre. A note from a saxophone, for instance, differs from a note of the same pitch and intensity produced by a violin or a xylophone; similarly vibrating reeds, columns of air, and strings all differ. Quality is dependent on the number and relative intensity of overtones produced by the vibrating body, and these, in turn, depend upon the nature of the vibrating body.
    Sounds may be generally characterized by pitch, loudness, and quality. Sound quality or timbre describes those characteristics of sound which allow the ear to distinguish sounds that have the same pitch and loudness. Timbre is a general term for the distinguishable characteristics of a tone.
    Hence, the quality or timbre characteristic of a sound helps you to identify your friend by his voice while sitting with others in a dark room.
  • Question 10
    1 / -0
    SONAR is
    Solution
    SONAR is the best example that illustrates how a ship on the ocean utilizes the reflecting properties of sound waves to determine the depth of the ocean. A sound wave is transmitted and bounces off the seabed. Because the speed of sound is known and the time lapse between sending and receiving the sound can be measured, the distance from the ship to the bottom of the ocean can be determined. This technique is called sonar (SOund Navigation And Ranging).
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