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jodygoodwin40
10.12.2019 •
Physics
Atransverse wave on a string is described with the wave function y(x, t) = (0.59 cm)sin[(1.80 m−1)x − (6.00 s−1)t]. (a) what is the wave velocity? (enter the magnitude in m/s.) m/s (b) what is the magnitude of the maximum velocity (in m/s) of the string perpendicular to the direction of the motion? m/s
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Ответ:
(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
Ответ: