Oscillation of a String: Amplitude, Speed & Node Distance

[matt62010]

A string oscillates according to the equation below.
y' = (0.50 cm) sin[(π/3 cm-1)x] cos[(45π s-1)t]

(a) What are the amplitude and speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation?
cm (amplitude)
cm/s (speed)
(b) What is the distance between nodes?
cm
(c) What is the speed of a particle of the string at the position x = 1.5 cm when t = 9/8 s?
cm/s

i have no clue how to do this problem. any help is greatly appreciated. thanks?

 

[Redbelly98]

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(from https://www.physicsforums.com/showthread.php?t=5374 )

 

[thomate1]

(a) The amplitude is 0.50 cm for both waves, as given by the coefficient in front of the sine function. The speed of the waves can be found by looking at the coefficient in front of the cosine function, which is 45π s^-1. This means that the waves are traveling at a speed of 45π cm/s in opposite directions.

(b) The distance between nodes can be found by dividing the wavelength by the number of nodes, which is given by the formula λ/n. In this case, the wavelength is 2π/(π/3 cm^-1) = 6 cm, and there are 2 nodes (one at the origin and one at the end of the string). Therefore, the distance between nodes is 6 cm/2 = 3 cm.

(c) To find the speed of a particle at a specific position and time, we can use the formula v = ωAcos(ωt - kx), where ω is the angular frequency (45π s^-1), A is the amplitude (0.50 cm), and k is the wave number (π/3 cm^-1). Plugging in the values, we get v = (45π s^-1)(0.50 cm)cos[(45π s^-1)(9/8 s) - (π/3 cm^-1)(1.5 cm)] = -20.62 cm/s. Note that the negative sign indicates that the particle is moving in the opposite direction of the wave's propagation.