Final answer:
The frequency of the wave is 20.0 Hz, the angular frequency is 40π rad/s, the wave number is π / 0.90 rad/m, the function y(x,t) that describes the wave is y(x,t) = 2.50 sin(πx/0.90 - 40πt), y(t) for a particle at the left end of the string is y(t) = 2.50 sin(-40πt), y(t) for a particle 1.35 m to the right of the origin is y(t) = 2.50 sin(π(1.35)/0.90 - 40πt), the maximum magnitude of transverse velocity of any particle of the string is 100π mm/s, the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s is y(x,t) = 2.50 sin(π(1.35)/0.90 - 40π(0.0625)), and the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s is calculated using the derivative of the equation from part (d).
Step-by-step explanation:
(a) The frequency of a wave can be calculated using the formula f = v / λ, where v is the wave speed and λ is the wavelength. In this case, the wave speed is 36.0 m/s and the wavelength is 1.80 m. Plugging in these values, we get f = 36.0 / 1.80 = 20.0 Hz.
(b) The angular frequency of a wave is given by the formula ω = 2πf, where f is the frequency. Substituting the value of f from part (a), we have ω = 2π(20.0) = 40π rad/s.
(c) The wave number of a wave is given by the formula k = 2π / λ, where λ is the wavelength. Plugging in the value of λ from the question, we have k = 2π / 1.80 = π / 0.90 rad/m.
(d) The function y(x,t) that describes the wave is given by the equation y(x,t) = A sin(kx - ωt), where A is the amplitude, k is the wave number, x is the position, t is the time, and ω is the angular frequency. Substituting the values of A, k, and ω from the question, we have y(x,t) = 2.50 sin(πx/0.90 - 40πt).
(e) To find y(t) for a particle at the left end of the string, we need to find the position x at the left end of the string. Since the origin is at the left end of the undisturbed string, the position x at the left end is 0. Plugging in x = 0 in the equation from part (d), we get y(t) = 2.50 sin(-40πt).
(f) To find y(t) for a particle 1.35 m to the right of the origin, we plug in x = 1.35 in the equation from part (d), we get y(t) = 2.50 sin(π(1.35)/0.90 - 40πt).
(g) The maximum magnitude of transverse velocity of any particle of the string can be found using the formula vmax = Aω, where A is the amplitude and ω is the angular frequency. Plugging in the values from the question, we get vmax = 2.50(40π) = 100π mm/s.
(h) To find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s, we plug in x = 1.35 and t = 0.0625 into the equation from part (d), we have y(x,t) = 2.50 sin(π(1.35)/0.90 - 40π(0.0625)).
(i) To find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s, we need to find the derivative of the equation from part (d) with respect to t. The derivative of sin(πx/0.90 - 40πt) with respect to t is -40πcos(πx/0.90 - 40πt). Plugging in x = 1.35 and t = 0.0625, we get the transverse velocity at that point in time.