Harmonic waves

1. Harmonic waves:
An example of harmonic waves are Tsunami and regular water waves which is
generated by wind.
These two harmonic waves differ in the following:
- The Wavelength of Tsunami waves are much larger than regular water waves. For
example, regular water waves can have a wavelength of 100 m but Tsunami waves can
have a wavelength over 500 km.
- The Amplitude of Tsunami waves are much larger than the Amplitude of regular
water waves and can exceed 30 m compared to 3 m in regular waves.
- The speed of Tsunami waves is much larger than the speed of regular water waves.
30 km/h is an average speed of Regular water waves but Tsunami water waves can
reach 700 km/h.
- Wave Period of Tsunami wave can reach an hour but regular water waves can have a
period of 1 sec.
As the regular water waves approach the coast, they lose their speed and energy and
therefore casuse no damage. Compared to Tsunami waves, as they approach the
coast, they will also lose some of their speed and energy but the loss is not significant.
Additionally, in deep water, tsunami waves take about 1 hour to complete a wave cycle
and its amplitude is sometimes not noticeable because of its long wave length. But as
the wave reaches the coast, the wavelength decreases along with its speed. The period
of the wave remains constant.The amplitude as a result increase and becomes
destructive as it hits the shores.
Equations:
Speed of a wave is defined as:
v = λ / T, where λ is wavelength and T is period of the wave
Tsunami waves follow shallow water equations and therefore the speed of tsunami
wave is:
v =√ 𝑔𝑑, where g is the force of gravity and equals 9.8 m/s2, d is the depth of the water.
Question:

2. A Tsunami of a wavelength of 200 km is generated 5000 km away from a shore line.
The depth of the water is assumed to be 4500 m. Calculate the following:
a. Speed of the wave.
b. The time it takes to hit the shore
c. The wave period and frequency
d. As the wave reaches the shores, the depth of the water becomes 15 m.
Calculate the new speed and wavelength.
Solution:
Part a:
Since v = √ 𝑔𝑑, then
v = √9.8 ∗ 4500 = 210 m/s
Part b:
To reach the shore which is 5000 km away, this wave needs (suppose constant speed):
Time = distance / velocity = 5000 * 1000 m / 210 = 23809 sec = 6.6 hours
Part c :
Since v = λ / T then T = λ / v, so period of the wave is:
T = 200 * 1000 / 210 = 962 sec = 16 min
It’s frequency f = 1 / T = 0.001 sec-1
Part d:
Velocity at the shore is calculated with the new depth:
v = √ 𝑔𝑑
v = √9.8 ∗ 15 = 12 m/s
For the wavelength, we assume the Wave period change is minimal, so:
V = λ / T or λ = v * T, so wave length of the new wave at the shore is :
λ = 12 * 962 / 1000 = 11.5 Km