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

Bailee Ayers

Physics

12 December, 20:37

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?

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Answers (1)

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Laila Blankenship · Answer 1

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