In the given transverse wave equation y = 0.05sin(5πt - 0.03πx):
Wave Amplitude (A): The wave amplitude (A) is the maximum displacement of the wave from its equilibrium position. In the given equation, the amplitude is 0.05. So, A = 0.05.
Frequency (f): The frequency (f) of the wave is the number of complete oscillations or cycles per unit time. In the given equation, the coefficient of "t" inside the sine function is 5π. This coefficient corresponds to the angular frequency (ω) of the wave. The frequency can be obtained by dividing ω by 2π:
ω = 5π f = ω / (2π) = 5π / (2π) = 5/2 ≈ 2.5 Hz
So, the frequency of the wave is approximately 2.5 Hz.
- Wave Propagation Speed (v): The wave propagation speed (v) represents how fast the wave is traveling through the medium. In the case of a transverse wave on a wire, the wave propagation speed (v) can be determined using the equation:
v = λ * f
where λ (lambda) is the wavelength of the wave.
To find the wavelength, we need to compare the given equation with the general form of a transverse wave:
y = A * sin(kx - ωt)
Comparing the two equations, we get:
kx - ωt = -0.03πx + 5πt
From this, we can extract the values of k (wave number) and ω (angular frequency):
k = -0.03π and ω = 5π
The wavelength (λ) is related to the wave number (k) as:
λ = 2π / |k|
So, in our case:
λ = 2π / |-0.03π| = 2π / 0.03π = 2 / 0.03 ≈ 66.67 units (where "units" depend on the unit of x).
Now, we can calculate the wave propagation speed (v):
v = λ * f = 66.67 * 2.5 ≈ 166.67 units per second (where "units" depend on the unit of x and t).
So, the wave propagation speed is approximately 166.67 units per second.