A large water tank has many corks bobbing on its surface. A simple wave propagat
ID: 2254993 • Letter: A
Question
A large water tank has many corks bobbing on its surface. A simple wave propagates from one end of the tank to the other. It is observed that the corks bob on the water with a maximum vertical speed of 3 cm/s and a maximum vertical acceleration of 2 cm/s2. (a) What is the amplitude of this water wave? (b) It is observed that a given row of corks at their maximum height (corresponding to a crest of the wave) are 3m away from the next crest of corks. What is the speed of propagation of the wave? the wave is a simple harmonic transverse wave.
Explanation / Answer
As we know that
Maximum Acceleration = w^2A
Maximum Velocity = wA
Divide both , we get
w = 2/3
Therefore
As
'
Maximum Velocity = wA
Therefore
A = 3/(2/3)
= 4.2 cm = 0.042 m
velocity = frequency*wavelength
= (w/2pi)*lambda
= (0.66667/(2pi))*(3000)
= 318.3 cm/sec
= 3.183 m/sec
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