Question

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 ripples being 0.21m/s wave length 15mm and amplitude 5 mm the maximum velocity of the piece of cork is

Answer 1

Maximum velocity of the piece of cork is v(max) = 0.44 m/s

Explanation:

We know that in wave motion, maximum particle velocity is given by.

max  = ωA = 2πfA

where ω = angular frequency

           f = linear frequency  

          A = amplitude

Given data:

wave velocity "u" =0.21 m/s

lambda = 15 mm = 0.015 m  

A = 5 mm=0.005 m

u=0.21 m/s

λ = 15 mm = 0.015 m

A = 5 mm = 0.005 m

Frequency = u / λ = 0.21/0.015 Hz

Therefore maximum particle (cork) velocity.

v(max)  = 2 × (22/7) × (0.21/0.015) × 0.005

           = 0.44 m/s  

Thus maximum velocity of the piece of cork is v(max) = 0.44 m/s

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