2. A certain string has a linear mass density of 0.25kg/m and is stretched with a tension of 25 N. One end is given a sinusoidal motion with frequency 5 Hz and amplitude 0.01 m. At t=0 the end has zero displacement and is moving in the +y direction. i.Find the wave speed.amplitide, angular frequency, period wavelength and wave number. [10m/s . 0.01 m. 31.4/s , 0.2 s, 2 m. 3.14/m] ii. Write a wave function describing the wave. [y=0.01sin(31.4t-3.14x)] iii Find the position of the point at x=0.25m at time t=0.1 second. [0.00707 m] iv. Find the transverse velocity of the point at x=0.25m at time. t=0.1 second. [-0.22 m/s]

Question

2. A certain string has a linear mass density of 0.25kg/m and is stretched with a tension of 25 N. One end is given a sinusoidal motion with frequency 5 Hz and amplitude 0.01 m. At t=0 the end has zero displacement and is moving in the +y direction. i.Find the wave speed.amplitide, angular frequency, period wavelength and wave number. [10m/s . 0.01 m. 31.4/s , 0.2 s, 2 m. 3.14/m] ii. Write a wave function describing the wave. [y=0.01sin(31.4t-3.14x)] iii Find the position of the point at x=0.25m at time t=0.1 second. [0.00707 m] iv. Find the transverse velocity of the point at x=0.25m at time. t=0.1 second. [-0.22 m/s]

Answer 1

i. The wave speed (v) can be calculated using the formula v = sqrt(T/μ), where T is the tension and μ is the linear mass density. Substituting the given values, we get v = sqrt(25/0.25) = 10 m/s.The amplitude (A) is given as 0.01 m.The angular frequency (ω) can be calculated using the formula ω = 2πf, where f is the frequency. Substituting the given value, we get ω = 2π(5) = 31.4 rad/s.The period (T) is the reciprocal of the frequency, so T = 1/f = 1/5 = 0.2 s.The wavelength (λ) can be calculated using the formula λ = v/ω. Substituting the calculated values, we get λ = 10/31.4 = 2 m.The wave number (k) can be calculated using the formula k = 2π/λ. Substituting the calculated value, we get k = 2π/2 = 3.14 rad/m.ii. The wave function can be written as y = A sin(ωt - kx), where A is the amplitude, ω is the angular frequency, t is the time, and k is the wave number. Substituting the given values, we get y = 0.01 sin(31.4t - 3.14x).iii. To find the position of the point at x = 0.25 m at time t = 0.1 s, we substitute these values into the wave function: y = 0.01 sin(31.4(0.1) - 3.14(0.25)) = 0.00707 m.iv. The transverse velocity (v) can be calculated using the formula v = ωy, where ω is the angular frequency and y is the displacement. Substituting the calculated values, we get v = 31.4(0.00707) = -0.22 m/s.

Вопрос

Решения

Ответ