Δλ = 3.103 m.
To solve this problem we have to know the speed of sound in both elements water and ice.
Element Speed of sound
Water (25°C) 1493 m/s
Ice 3200 m/s
The velocity of a wave is given by the equation v = λf, where λ is the wavelength, and f is the frecuency of the wave.
In order to calculate the wavelength we have to clear λ in the equation v = λf, resulting:
λ = v/f
Calculating the wavelength in both elements:
λ(water) = 1493 m/s / 550 Hz = 2.715 m
λ(ice) = 3200 m/s / 550 Hz = 5.818 m
So, the difference in wavelength between the wave produced in ice and the wave produced in water is:
Δλ = λ(water) - λ(ice) = 5.818 m - 2.715 m
Δλ = 3.103 m
B. 3.1 m is the correct answer on edge just took the test
V = ∧ f
Solving for wavelength on ice:
∧ = v / f
∧ = 3200 / 550
∧ = 5.82
Solving for wavelength in water:
∧ = 1500 / 550
∧ = 2.73
The difference is:
Difference = 5.82 - 2.73
Difference = 3.1
The answer is the letter "B. 3.1 m".
You forgot to mention speed of sound in ice and water.
speed of sound in water = 1500m/s
speed of sound in ice = 3200m/s
Frequency f = 550Hz.
Velocity of sound is given by
v = λf
λ = v/f
λ1 = 1500/550
λ1 = 2.73 m
λ2 = 3200/550
λ2 = 5.82 m
Difference in wavelength
= λ2 - λ1
= 3.09 m
Difference in wavelength for wave produced in ice and water is 3.09 m.
Hope this helps you...