Respuesta :
Answer:
a. frequency = 25 Hz, period = 0.04 s , wave number = 19.63 rad/m
b. y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]
c. 0.0496 m
d. 0.03 s
Explanation:
a. Frequency, f = v/λ where v = wave speed = 8.00 m/s and λ = wavelength = 0.320 m
f = v/λ = 8.00 m/s ÷ 0.320 m = 25 Hz
Period, T = 1/f = 1/25 = 0.04 s
Wave number k = 2π/λ = 2π/0.320 m = 19.63 rad-m⁻¹
b. Using y = Asin(kx - ωt) the equation of a wave
where y = displacement of the wave, A = amplitude of wave = 0.0700 m and ω = angular speed of wave = 2π/T = 2π/0.04 s = 157.08 rad/s
Substituting the variables into y, we have
y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]
c. When x = 0.360 m and t = 0.150 s, we substitute these into y to obtain
y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]
y = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s × 0.150 s)]
y = (0.0700 m)sin[(7.0668 rad) - (23.562 rad)]
y = (0.0700 m)sin[-16.4952 rad]
y = (0.0700 m) × 0.7084
y = 0.0496 m
d. For the particle at x = 0.360 m to reach its next maximum displacement, y = 0.0700 m at time t. So,
y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]
0.0700 m = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s)t]
0.0700 m = (0.0700 m)sin[(7.0668 rad - (157.08 rad/s)t]
Dividing through by 0.0700 m, we have
1 = sin[(7.0668 rad - (157.08 rad/s)t]
sin⁻¹(1) = 7.0668 rad - (157.08 rad/s)t
π/2 = 7.0668 rad - (157.08 rad/s)t
π/2 - 7.0668 rad = - (157.08 rad/s)t
-5.496 rad = - (157.08 rad/s)t
t = -5.496 rad/(-157.08 rad/s) = 0.03 s
The frequency of the wave is 25 Hz, the period is 0.04 s, and the wave number of the given wave is 19.63 rad/m.
(A) Frequency of the wave,
[tex]\bold {f = \dfrac v{\lambda}}[/tex]
where
v - speed = 8.00 m/s
λ - wavelength = 0.320 m
[tex]\bold {f = \dfrac {8.0 m/s }{0.32\ m} = 25\ Hz}[/tex]
Period of the wave,
[tex]\bold {T = \dfrac 1f = \dfrac 1{25 } = 0.04\ s }[/tex]
Wave number of the wave,
[tex]\bold {k = \dfrac {2\pi }{\lambda} = \dfrac {2\pi }{0.320} = 19.63\ rad\ m^{-1}}[/tex]
(B)
Wave function equation of a wave,
y = A sin (kx - ωt)
where
y - displacement of the wave
A = amplitude = 0.0700 m
ω = angular speed = 2π/T = 2π/0.04 s = 157.08 rad/s
Put the values in the formula,
y = (0.0700 m)sin[(19.63 rad/m)x - (157.08 rad/s)t]
(C) When x = 0.360 m and t = 0.150 s,
y = (0.0700 m)sin[(19.63 rad/m × 0.360 m) - (157.08 rad/s × 0.150 s)]
y = 0.0496 m
Therefore, the frequency of the wave is 25 Hz, the period is 0.04 s, and the wave number of the given wave is 19.63 rad/m.
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