Question

A fizzy drink can bobs up and down in water due to waves travelling across the surface. The waves travel at 0.7ms-1 and have a wavelength of 0.5m and an amplitude of 0.1m. Calculate the maximum velocity of the can?

Answer 1

Question 5177

A fizzy drink can bobs up and down in water due to waves travelling across the surface. The waves travel at 0.7ms-1 and have a wavelength of 0.5m and an amplitude of 0.1m. Calculate the maximum velocity of the can?

Solution:

Lets write the equation for monochromatic wave, with given amplitude, frequency and wavelength:

$$y (x, t) = A \sin (k x + \omega t) = A \sin \left(\frac {2 \pi}{\lambda} x + 2 \pi v t\right)$$

We can find the $\nu_{y}$ , by differentiating (1):

$$v _ {y} = \frac {\partial y}{\partial t} = A \cdot 2 \pi v \cdot \cos \left(\frac {2 \pi}{\lambda} x + 2 \pi v t\right)$$

Hence, the maximum $\nu_{y}$ is

$$v _ {y; \max } = A \cdot 2 \pi v$$

Also, we have a phase velocity for a monochromatic wave:

$$v _ {p} = \frac {\omega}{k} = \lambda \cdot v$$

So, the maximum speed for the can is given by:

$$v _ {\max } = \sqrt {\left(v _ {y ; \max } ^ {2} + v _ {p} ^ {2}\right)} = v \sqrt {\left(4 \pi^ {2} A ^ {2} + \lambda^ {2}\right)}$$

Using numeric values given, and putting it into (5), we obtain:

$$v _ {\max } = 5. 6 2 \cdot 1 0 ^ {- 4} \frac {m}{s}$$