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The equation describing a transverse wave on a string is y(x,t) = (1.50 mm) sin[(157 s-¹)t − (41.9 m-¹) x]. Find (a) the wavelength, frequency, and amplitude of this wave, (b) the speed and direction of motion of the wave, and (c) the transverse displacement of a point on the string when t=0.100 s and at a position x=0.135 m.

User Hytromo
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Final answer:

The transverse wave on a string has an amplitude of 1.50 mm, a wavelength of approximately 0.150 m, and a frequency of about 25 s⁻¹. The speed of the wave is roughly 3.75 m/s, moving in the positive x-direction. At t=0.100 s and x=0.135 m, the transverse displacement of a string point is approximately 1.50 mm.

Step-by-step explanation:

The equation y(x,t) = (1.50 mm) sin[(157 s⁻¹)t − (41.9 m⁻¹) x] describes a transverse wave on a string. To determine the characteristics of this wave, we examine the general form of a wave function, A sin(kx − ωt), where A is the amplitude, k is the wave number, and ω is the angular frequency.

Wave Characteristics:

(a) From the given equation, the amplitude A is 1.50 mm (0.0015 m), since this is the coefficient in front of the sine function. The angular frequency ω is 157 s⁻¹, and the wave number k is 41.9 m⁻¹. The wavelength λ can be found using the relationship k = 2π/λ, resulting in a wavelength of λ = 2π/k ≈ 0.150 m. The frequency f is ω/2π ≈ 25 s⁻¹.

(b) The wave speed v can be found using the relation v = fλ ≈ 3.75 m/s. The negative sign in the wave function denotes that the wave is moving in the positive x-direction.

(c) The transverse displacement for t=0.100 s and x=0.135 m is obtained by plugging these values into the wave equation, resulting in y(0.135, 0.100) ≈ 1.50 mm sin[(15.7) - (5.654)] ≈ 1.50 mm sin(10.046) ≈ 1.50 mm.

User Josh Reich
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