Answer:
The sound of a whale is especially low frequency, but there is a species of whale the Whalien whose frequency is 52 Hz, if the propagation speed of the wave is 1400m / s What will be its period in the water and the air? And what will be the wavelength in each medium? Remember that the propagation speed in air is 340m / s
Explanation:
From wave equation, the speed, wavelength and frequency is related using
V = fλ
Where
V is the speed
f is the frequency
And λ is the wavelength
So,
The frequency of the whale is
f = 52Hz
The speed in water is V_w = 1400m/s
The speed in air is V_a = 340m/s
We want to find the period in each medium, the period is related to the frequency and since the frequency is constant.
Then, period in equal in both medium
T = 1 / f
T_w = T_a = 1 / f
T = 1 / 52
T = 0.0192 seconds
We want to find the wavelength in each medium
For water,
V = fλ
V_w = f × λ_w
Then,
λ_w = V_w / f.
λ_w = 1400 / 52 = 26.92 m
The wavelength in water is 26.92m
Now, in air
V = fλ
V_a = f × λ_a
Then,
λ_a = V_a / f.
λ_a = 340 / 52 = 6.54 m
The wavelength in air is 6.54 m
In Spanish
De la ecuación de onda, la velocidad, la longitud de onda y la frecuencia se relacionan usando
V = fλ
Dónde
V es la velocidad
f es la frecuencia
Y λ es la longitud de onda
Entonces,
La frecuencia de la ballena es
f = 52Hz
La velocidad en el agua es V_w = 1400m / s
La velocidad en el aire es V_a = 340m / s
Queremos encontrar el período en cada medio, el período está relacionado con la frecuencia y dado que la frecuencia es constante.
Luego, período igual en ambos medios
T = 1 / f
T_w = T_a = 1 / f
T = 1/52
T = 0.0192 segundos
Queremos encontrar la longitud de onda en cada medio
Para agua,
V = fλ
V_w = f × λ_w
Entonces,
λ_w = V_w / f.
λ_w = 1400/52 = 26,92 m
La longitud de onda en el agua es de 26,92 m.
Ahora en el aire
V = fλ
V_a = f × λ_a
Entonces,
λ_a = V_a / f.
λ_a = 340/52 = 6,54 m
La longitud de onda en el aire es de 6.54 m.
