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Sound Test - 21

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Sound Test - 21
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  • Question 1
    1 / -0
    The full name of 'SONAR' is.
    Solution
    Answer is C.
    The acronym SONAR stands for SOund Navigation And Ranging. Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. 
  • Question 2
    1 / -0
    Sounds having frequency more than 20,000 Hz are called -
    Solution
    Sounds having frequency more than 20,000 Hz are called Ultrasonic sound. As this is above the normal hearing range for humans, we cannot hear ultrasonic sound.
  • Question 3
    1 / -0
    If a vibrator strikes the water 50 times in 5 second, then the frequency of wave is________.
    Solution
    In  $$5 seconds$$ it vibrates $$50 times$$
    In one sec it will vibrate $$ \dfrac{50}{5} = 10 times$$
    And the number of vibrations in one second is nothing but frequency defined as number of oscillations in one second .
    $$freq = \dfrac{10 times}{1s} = 10 Hz.$$
    Where $$1Hz$$ = one oscillation per second.
  • Question 4
    1 / -0
    A sound travels at a speed of 798 m/s. If its wavelength is 3m. What is the frequency of its wave. Will be be audible or not :
    Solution
    Given - $$v=798m/s  ,  \lambda=3m$$ ,
    We have  $$v=n\lambda$$ , where $$n=$$ frequency of wave ,
    Therefore ,  $$n=v/\lambda=798/3=266Hz$$ .
    The audible frequency range for the human ear is from $$20Hz$$ to $$20,000Hz$$.  So the frequency of the given sound - $$266Hz$$ lies in the audible range.

  • Question 5
    1 / -0
    Select the odd one out (loudness, pitch, quality, brightness)
    Solution
    All are characteristics of sound except brightness
  • Question 6
    1 / -0
    The frequency of sound is 500 Hz. How many times it vibrate in a minute :
    Solution
      Frequency of vibration is defined as number of vibrations produced per second .
       Given , frequency of sound 500Hz , 
       it means that in $$1$$ second , number of  vibrations  produced $$=500$$
       therefore , in 60 second (1 minute) , number of vibrations produced $$500\times60/1=30,000$$      
  • Question 7
    1 / -0
    Compressions and rarefactions are seen in
    Solution
    Longitudinal waves propagate in a medium in the form of compressions and rarefactions . When pressure becomes high , medium particles come closer and a compression is formed and due to low pressure , rarefaction is formed .
    Transverse waves propagate in the form of crests and troughs .
  • Question 8
    1 / -0
    A tuning fork produces 20 vibrations per second. What is the frequency of the tuning fork:
    Solution
    Frequency is defined as the number of vibrations produced per second, as the given tuning fork produces 20 vibrations per second, therefore its frequency is $$20Hz$$.
  • Question 9
    1 / -0
    A source produces 60 rarefactions and 60 compressions in 0.6 s. Distance between a compression and the next (consecutive) rarefaction is 150 cm. Find the frequency of the wave :
    Solution
    Frequency of wave is defined as the number of waves produced in 1 second. A longitudinal wave is formed by one compression and one rarefaction, we have 60 compressions and 60 rarefactions in $$0.6s$$ i.e. 60 waves produced in 0.6s,
    In $$0.6s $$ the number of wave produced $$=60$$
    In $$1s$$ the number of wave produced $$=\dfrac{60}{0.6}=100Hz$$ 
    Therefore , frequency $$f=100Hz$$

    Here, the distance given between a compression and the next (consecutive) rarefaction is of no use. It is wavelength. We need only number of waves and time to calculate frequency.
  • Question 10
    1 / -0
    A source of wave produces 1 crest and 1 trough in 16 s. Find the frequency of the wave.
    Solution
    Let's find out the time period of the wave-
    One cycle of the wave consists 1 crest and 1 trough.
    So the frequency of the wave
    $$T = 16\ s$$
    So the frequency of the wave

    $$\text{ Frequency } = \dfrac{ 1 }{ \text{Time period} }$$

    $$\Rightarrow \text{ Frequency } = \dfrac{ 1 }{16 }$$

    $$\Rightarrow \text{ Frequency } = 0.0625\ Hz$$
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