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Sound Test - 31

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Sound Test - 31
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which of the following is not a longitudinal wave?
    Solution
    In case of light, the waveform changes perpendicular to the relative motion. So, it is a type of transverse wave, which causes the medium to vibrate perpendicular to the wave energy .
  • Question 2
    1 / -0
    Which of the following is mechanical wave?
    Solution
    Mechanical waves needs a media through which they can transmit, and media must have elastisity. Sound waves are mechanical waves, which travels by the particles/atoms or molecules of media.
  • Question 3
    1 / -0
    The persistence of sound in a hall is called
    Solution
    The persistence of sound in a hall is called reverberation.
    Option 'A' is correct.

  • Question 4
    1 / -0
    How many types of mechanical waves exist in physics?
    Solution
    There are two basic types of mechanical waves: longitudinal waves and transverse waves.
  • Question 5
    1 / -0
    A stone is dropped into a well and the sound of impact of stone with the water is heard after $$2.056\ \text{s}$$ of the release of stone from the top. If the acceleration due to gravity is $$980\ \text{cm}s^{-2}$$ and the velocity of sound in air is $$350\ \text{ms}^{-1}$$, then the depth of the well is
    Solution
    Time taken by the stone to reach the water surface

    $$t_1 = \sqrt{\dfrac{2h}{g}}$$, where h is the height of the well.

    Time taken by the sound to reach the observer

    $$t_2 = \dfrac{h}{350}$$

    Total time

    $$t= t_1 +t_2 = \sqrt{\dfrac{2h}{g}} + \dfrac{h}{350}= \sqrt{\dfrac{h}{4.9}} + \dfrac{h}{350} = 2.056$$

    Solving the above Eqn. we get

    $$h = 19.6\ \text{m}$$
  • Question 6
    1 / -0
    A man standing unsymmetrically between two parallel cliffs, claps his hand and starts hearing a series of echoes at intervals of 1 s. If the speed of sound in air is 340 m$$s^{-1}$$, then the distance between the two prallel cliffs, is
    Solution

    Let the man $$M$$ be at distance $$x$$ from hill $$H_1$$ and $$y$$ from hill $$H_2$$ as shown in figure. Let $$y > x$$.
    The time interval between the original sound and echos from $$H_1$$ and $$H_2$$ will be respectively
    $$t_1 = \dfrac{2x}{v}$$ and $$t_2 = \dfrac{2y}{v}$$, where $$v$$ is the velocity of sound.
    The distance between the hills is 
    $$x + y = \dfrac{v}{2} [t_1 + t_2] = \dfrac{340}{2}[1 + 2] = 510 $$ 

  • Question 7
    1 / -0
    Surface waves strike the rock with their crests $$160\ m$$ apart. The velocity of the wave is $$40\ ms^{-1}$$, The time interval between two crests striking the rock is
    Solution
    Given,
    The distance between crest is called wavelength.
    $$ \lambda = 160\ \text{m}$$  
    $$\text{velocity(v)} = 40\ ms^{-1} $$
    so, 
    $$\text{time(t)} = \dfrac { \lambda }{ v }$$

    $$\Rightarrow t = \dfrac { 160 }{ 40 }$$

    $$\Rightarrow t = 4\ \text{s}$$
  • Question 8
    1 / -0
    Which type of wave is produced in the stem of tuning fork?
    Solution
    When tuning fork is sounded by striking its one end on rubber pad then the prongs vibrate in and out and stem vibrate up and down. Hence, vibration of prongs are transverse and those of stem are longitudinal. 
  • Question 9
    1 / -0
    A wave passes from one medium to another medium. Which of the following quantity will change ?
    Solution
    Sound is propagated in the form of longitudinal waves. When sound travels from one medium to another, both its velocity and wavelength undergo changes. The velocity of sound in a given medium is obtained by the equation $$v=f \lambda$$, where $$f$$ is the frequency of sound and $$\lambda$$ is its wavelength in that medium. 

    Velocity of sound is directly proportional to the wavelength. Thus, if the velocity of sound doubles when it travels from one medium to another, its wavelength also doubles. The frequency of sound depends upon the source of sound, not the medium of propagation. Hence, it does not change. 
    Same argument is valid for all waves as well.
  • Question 10
    1 / -0
    The wavelength of waves produced on the surface of the water is $$20\ cm$$. If the wave velocity is $$24\ ms^{-1}$$, calculate the number of waves produced in one second.
    Solution
    In the question, the wavelength $$\left( \lambda  \right) $$ is given as $$20\ cm$$, that is, $$ 0.20\ m$$ and the wave velocity v is given as $$24$$ $$m/s$$.

    The frequency $$\nu $$, velocity $$v$$ and wavelength $$\lambda $$ relation is given as:
    $$\nu =\dfrac { v }{ \lambda  } $$

    Number of waves produced in one second is called as frequency, i.e. $$\nu =\dfrac { v }{ \lambda  } $$ $$=\nu =\dfrac { 24m/s }{ 0.20m } $$ $$= 120$$ waves per second.
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