A piece of cork is floating on water in a small tank. The cork oscillates up and down vertically when small ripples pass over the surface of water. The velocity of the ripples being $0.21 m s^{- 1}$ , wave length 15 mm and amplitude 5 mm, the maximum velocity of the piece of cork is $(π = \frac{22}{7})$

0.44 m s^{- 1}

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0.24 m s^{- 1}

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2.4 m s^{- 1}

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4.4 m s^{- 1}

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Solution

The correct option is A $0.44 m s^{- 1}$
We know that in wave motion , maximum particle velocity is given by ,
$v_{m a x} = ω A = 2 π f A$ ,
where $ω =$ angular frequency ,
$f =$ linear frequency ,
$A =$ amplitude ,
given - wave velocity , $u = 0.21 m / s, λ = 15 m m = 0.015 m, A = 5 m m = 0.005 m$ ,
by , $f = u / λ = 0.21 / 0.015$ Hz ,
therefore maximum particle (cork) velocity ,
$v_{m a x} = 2 \times (22 / 7) \times (0.21 / 0.015) \times 0.005 = 0.44 m / s$

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A piece of cork id floating on water in a small tank. The cork oscillates up and down vertically when small ripples pass over the surface of water. The velocity of the ripples being 0.21 ms^-1, wave length 15 mm and amplitude 5 mm, the maximum velocity of the piece of cork is

Q. A piece of cork is floating in water in a small tank. The cork oscillates up and down vertically when small ripples pass over the surface of the water. The velocity of the ripples being

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, wavelength 15 mm and amplitude 5 mm, find the maximum velocity of the piece of cork

(π = \frac{22}{7})

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A piece of cork is floating on water in a small tank. The cork oscillates up and down vertically when small ripples pass over the surface of water. The velocity of the ripples being 0.21 ms−1, wave length 15 mm and amplitude 5 mm, the maximum velocity of the piece of cork is (π=227)