Waves

  • Question 1
    CBSEENPH11020367

    Three sound waves of equal amplitudes have frequencies (ν – 1), ν, (ν + 1). They superpose to give beats. The number of beats produced per second will be.

    • 4

    • 3

    • 2

    • 1

    Solution

    C.

    2

    p1 =po sin 2π(x- 1)t
    p2 =po sin 2π(x)t
    p3 =po sin 2π(x+ 1)t
    p=p1+p3+p2
    =posin2π(x-1)t +po sin 2π(x+1)t +po sin 2π(x)t
    = 2posin2πxtcos2πt +posin2πxt
    = 2posin2πxt[2cos πt +1]
    ⇒ fbeat = 2

    Question 2
    CBSEENPH11020384

    A wave travelling along the x-axis is described by the equation y(x, t) = 0.005 cos (αx −βt). If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then α and β in appropriate units are 

    • α = 25.00 π, β = π

    • α = 0.08/ π,  2.0/π

    • α = 0.04 / π , β =1.0/π

    • α = 12.50 , β =π/ 2.0

    Solution

    A.

    α = 25.00 π, β = π

    y = 0.005 cos (αx − βt)
    comparing the equation with the standard form,
    straight y space equals space straight A space cos space open square brackets open parentheses straight x over straight lambda minus straight t over straight T close parentheses 2 straight pi close square brackets
2 straight pi divided by straight lambda space equals space straight alpha space and space 2 straight pi divided by straight T space equals space straight beta
straight alpha space equals space 2 straight pi divided by 0.08
space equals space 25.00 space straight pi
straight beta space equals space straight pi

    Question 3
    CBSEENPH11020407

    A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms−1. The velocity of sound in air is 300 ms−1. If the person can hear frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear the whistle is

    • 15 square root of 2 space ms to the power of negative 1 end exponent
    • 15 divided by square root of 2 space ms to the power of negative 1 end exponent
    • 15 ms-1

    • 30 ms-1

    Solution

    C.

    15 ms-1

    straight f subscript app space equals space fraction numerator straight f left parenthesis 300 right parenthesis over denominator 300 minus straight v end fraction
rightwards double arrow space straight v space equals space 15 space straight m divided by straight s
    Question 4
    CBSEENPH11020457

    When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per seconds are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?

    • 200 Hz

    • 202 Hz

    • 196 Hz

    • 204 HN

    Solution

    C.

    196 Hz

    |f1−f2| =4 Since mass of second tuning fork increases so f2 decrease and beats increase so f1>f2 ⇒ f2=f1−4 = 196

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