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Sound Test - 24

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Sound Test - 24
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  • Question 1
    1 / -0
    Which of the above two graphs (a) and (b) representing the human voice is likely to be the male voice ? Give a reason for your answer.

    Solution
    The two vocal properties are intensity (loudness) and frequency (pitch). Pitch of a man's voice fall under low frequency, whereas woman's voice is of the high pitch type. Pitch and intensity are proportional to each other. As seen in the graph given, we can compare the intensities of male and female voice. Female's voice has more frequency components marked as 'b' compared to men which is marked as 'a'. Women speak at one octave higher than men. 
    Hence, the wave 'a' represents a male voice.
  • Question 2
    1 / -0
    The loudness of sound depends on:
    Solution
    Some of the sounds appear loud whereas others appear to be faint in the ear. The faint sound has a small amplitude whereas the loud sound has a large amplitude.Thus,the loudness or softness of a sound is determined basically by the amplitude of the wave. 
  • Question 3
    1 / -0
    The loudness of sound is determined by the
    Solution
    The amplitude of sound determines the loudness of  sound. A sound that is faint will have a small amplitude while that of a loud sound will have larger amplitude.
  • Question 4
    1 / -0
    When we change feeble sound to loud sound we increase its :
    Solution
    The amplitude is the measure of loudness of a sound. When our music system is having a feeble sound by increasing the amplitude, the sound now becomes louder.
  • Question 5
    1 / -0
    In the bell jar experiment, as air is removed from the jar :
  • Question 6
    1 / -0

    Time period of a sound wave is $$10\ ms$$. What is the frequency of the sound wave?
    Solution
    Given
    $$T= 10\ ms = \dfrac{1}{1000} = 10^{-2}\ s$$
    We know
    $$f= \dfrac{1}{T}$$
    $$f= \dfrac{1}{10^{-2}} = 100\ Hz$$
  • Question 7
    1 / -0
    The frequency of sound waves can be expressed in
    Solution
    The frequency means the number of full cycle of the wave per second. So all the options can be used to express the frequency. They all have same dimensions $$[ T ]^{-1}$$.
  • Question 8
    1 / -0
    Which of the following cannot travel through vacuum?
    Solution
    Light waves, heat waves, X-rays are examples of electromagnetic waves, which do not require any medium for their propagation.
    Sound waves cannot travel through a vacuum. It needs a material medium for its propagation. Sound waves can propagate only when the vibrational energy of the particles of the medium is travelling from one point to another.
  • Question 9
    1 / -0
    If a wave completes 20 vibrations in 2.5s, then its frequency is
    Solution
    Frequency is the number of vibration per second.
    So in this case the frequency of the wave is:
    $$f=\dfrac{20}{2.5} Hz$$ $$=8\ Hz$$
  • Question 10
    1 / -0
    Hertz is the unit of:
    Solution
    The number of cycles per unit time is known as frequency. The SI unit of frequency is Hertz $$(Hz)$$.
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