Question

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a wave laps against the pier every 5.0 s, what are (a) the frequency and (b) the speed of the wave?

Answer 1

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$.

Part(b): The speed of the wave is $\bf{0.5~m/s}$.

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$.

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$.

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$.

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$