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Waves Test - 3

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Waves Test - 3
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  • Question 1
    1 / -0

    The fundamental frequency of a closed organ pipe is 660 Hz. If the second overtone frequency of an open organ pipe is equal to the first overtone frequency of the closed organ pipe, then what is the length of the open organ pipe? (Speed of sound in air = 330 m/s)

    Solution

    First overtone frequency of the closed organ pipe = 3 × fundamental frequency = 3 × 660 = 1980 Hz

    Second overtone frequency of the open organ pipe = 3v/2L; where v = speed of sound in air and L= length of open organ pipe.
    According to question,
    Second overtone frequency of the open organ pipe = First overtone frequency of the closed organ pipe

    Therefore, 3v/2L = 1980 Hz

    On putting the value of v = 330 m/s, we get L = 25 cm
    Hence, the length of the open organ pipe is 25 cm.

  • Question 2
    1 / -0

    When there is superposition of the two waves of nearly equal frequencies, then beat frequency is defined as

    Solution

    When there is superposition of two waves of nearly equal frequencies, then beat frequency is defined as the number of times the resultant intensity becomes maximum or minimum in one second.

  • Question 3
    1 / -0

    In the following question, a statement of assertion followed by a statement of reason is given. Read both the statements carefully and choose the correct option.

    Assertion: Amplitude of a cylindrical progressive wave is a/ √r.

    Reason: The energy per unit area of the wave decreases as r increases.

    Solution

    Since energy is conserved, the energy per unit area of the cylindrical wave decreases as r increases. (Area of a cylindrical wave = 2πrh)
    So, the intensity of the wave will be inversely proportional to r. Hence, amplitude = a/√r, as amplitude is directly proportional to the square root of the intensity of wave.
    Therefore, the equation of the cylindrical progressive wave will be a/√r sin(ωt - kr).

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