Answer:
(a) wave velocity = 3.33 m/s².
(b) magnitude of the maximum velocity = 3.54 m/s²
Explanation:
The general equation of a traveling wave is given,
y = Asin(ωt - kx)...................................(equation 1)
Where A = Amplitude of the wave (m)
ω = Angular frequency (s⁻¹)
k = Angular wave number (m⁻¹)
(a).
From the expression above, v = ω/k
Given : y = (0.59)sin[(1.80)x − (6.00)t ..................(equation 2)
Comparing Equation 1 and equation 2
1.8x = -kx
∴ k = -1.8 m⁻¹
And -6.00t = ωt
∴ ω = -6.00 s⁻¹
∴ v = - 6.00/-1.8 = 3.33 m/s².
wave velocity = 3.33 m/s².
(b).
We differentiate (equation 2) with respect to time (t) to get an expression for the transverse speed of the wave.
∴ dy/dt = d{(0.59)sin[(1.80)x − (6.00)t]/dt
dy/dt = 0.59(-6.00)cos(1.8x - 6.00t)
dy/dx = 0.59(-6.00)cos(1.8x - 6.00t)
The magnitude of the maximum velocity = The absolute value of the coefficient of the cosine function.
Vmax = 0.59 × 6.00 = 3.54 m/s²
∴ magnitude of the maximum velocity = 3.54 m/s²
