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A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,

(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.

User JustTB
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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.

User Barrrdi
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