Transverse waves on a string have wave speed v=8.00 m/s, amplitude A=0.0700m, and direction, and at t=0 the x-0 end of the wavelength -0.320m. The waves travel in the -x string has its maximum upward displacement. a) Find the frequency, period and wave number of these waves b) Write a wave function describing the wave c) Find the transverse displacement of a particle at x=0.360m at time t=0.150 d) How much time must elapse from the instant in part (c) until the particle at x-0.360 m next has maximum upward displacement?
Question
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