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Sound Test - 54

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Sound Test - 54
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Weekly Quiz Competition
  • Question 1
    1 / -0
    What is the frequency of a vibrating body whose time-period is 0.05 second?
    Solution
    Frequency $$f$$ and time period $$t$$ are reciprocal of each other.
     $$f = \dfrac{1}{t} = \dfrac{1}{0.05} = \dfrac{100}{5} = 20 Hz$$
  • Question 2
    1 / -0
    In the bell jar experiment, after starting the vacuum pump the sound gradually become fainter and finally we hear a feeble sound. Why?
    Solution
    Sound needs a medium to travel. The vacuum pump pumps out the air, and thus the medium through which the sound travels is pumped out and so the sounds get fainter.
  • Question 3
    1 / -0
    The loudness or softness of a sound is determined basically by ________.
    Solution
    Loudness or softness of a sound is determined basically by its amplitude. The amplitude of the sound wave depends upon the force with which an object is made to vibrate. If we strike a table lightly, we hear a soft sound because we produce a sound wave of less energy (amplitude). If we hit the table hard we hear a loud sound.Loud sound can travel a larger distance as it is associated with higher energy.
  • Question 4
    1 / -0
    If the wavelength of the wave is $$x$$. Then the distance between compression and the adjoining rarefaction is:
    Solution
    A single wave consists of compression and rarefaction. So the distance between them will be $$\dfrac{x}{2}$$.
  • Question 5
    1 / -0

    Fix up one end of a slinky to a hook and hold it horizontally. Vibrate the other end to and fro along the length of the spring. If you look at the spring carefully , you will see some parts of the spring are pushed closer , while some other parts are pulled apart.Then which of the following statements are correct about above experiment?

    Solution
    Consider a vibrating tuning fork.
    A vibrating tuning fork is capable of creating a longitudinal wave. As the prongs of the fork vibrate back and forth, they push on neighboring air particles. The forward motion of a prong pushes air molecules horizontally to the right and the backward retraction of the prong creates a low-pressure area allowing the air particles to move back to the left.So the air molecules also move in longitudinal manner.

    Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart. These regions are known as compressions and rarefactions respectively. The compressions are regions of high air pressure (particles are closer) while the rarefactions are regions of low air pressure(particles are farther). 
    The slinky experiment is just like the sound waves.The spring behave as the medium so spring body parts show longitudinal vibrations like sound so some part of slinky is in compression (spring is compressed) while some part is in rarefaction (spring is pulled apart) .
  • Question 6
    1 / -0
    How ultra sound is used to clean hard to reach places?
    Solution
    Ultrasonic waves are used for cleaning hard to reach places such as spiral tube, odd shaped machine parts etc. The object to be cleaned is kept in the cleaning solution and the solution is subjected to the ultrasonic waves. The ultrasonic waves stir up the dust/dirt particles, causing the particles to get detached from the object and hence the object is thoroughly cleaned.
  • Question 7
    1 / -0
    On stretching a slinky, the regions where the coils become closer are called _____ and the regions where coils are further apart are called ____.
    Solution
    Compressions are the area of higher density or larger number of turns per unit length are present in these regions compared to other regions. The portions which are pulled apart have fewer turns per unit length than the normal spring are called rarefactions.
  • Question 8
    1 / -0
    A boy standing at some distance from a cliff claps his hands and heard an echo $$3s$$ later. What is the distance of the cliff from the boy, if the speed of the sound is $$340 m{s}^{-1}$$.
    Solution
    Given:
    The speed of the sound $$v=340 m{s}^{-1}$$
    Time taken by the echo to reach the boy is $$t=3\ s$$
    If $$d$$ is the distance between the boy and the cliff then in $$3\ s$$ the sound has to travel a distance of $$2d$$.

    Using the equation $$S=vt$$,
    $$2d=vt= 340 \times 3$$ 
    $$\Rightarrow d=\dfrac{340\times 3}{2}$$
    $$\Rightarrow d=170\times 3 = 510\ m$$

    So, the distance between the boy and the cliff is $$510\ m$$.
  • Question 9
    1 / -0
    A sonar echo takes 8.8s to return from a shark. How far away is the shark?
    Speed of the sound in sea water at $${25}^{\circ}C$$ is $$1533\ m{s}^{-1}$$
    Solution
    Total time is taken by the signal 
    $$T= 8.8\ s$$
    Time taken by the signal to reach the shark
    $$t = \dfrac{T}{2}$$
    $$\Rightarrow t = \dfrac {8.8}{2}$$
    $$\Rightarrow t=4.4\ s$$
    Speed of sound in sea water at $$25^{\circ}C= 1533\ ms^{-1}$$
    Distance of shark, 
    $$d = vt$$
    $$\Rightarrow d = 1533 \times 4.4$$
    $$\Rightarrow d = 13490\ m$$
  • Question 10
    1 / -0
    If the ceiling of a cinema hall is made flat, what will happen?
    Solution
    If the ceiling of the cinema hall is flat then the sound will undergo regular reflection and only the audience sitting at a place will be able to hear the sound.  
    The ceiling in a cinema hall is made curve so that the sound of music, dialogue, etc must reach evenly to all the audience. So that the sound after reflection from the curved surface reaches all corners of the hall evenly.

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