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A string with linear density 4.0 g/m is lies along the positive x-axis with tension 40 N. One end of the string is fixed to a wall, and the other is attached to an actuator at x=0 m that oscillates up and down with a frequency of 50 Hz. The actuator moves with a maximum displacement of 2.5 mm, and at t=0 s is at the center of it's height, moving downwards: 1. What is the wave speed on the string, and the wavelength? 2. What is the amplitude of the wave, and its phase constant? 3. Write the displacement equation D(x,t) of the traveling wave. 4. What is the string's displacement at x=0.50 m and t=10 ms?

User Atzoya
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To answer these questions, we must apply the formulas related to waves and their properties.

1. The wave speed, v, can be calculated using the formula v = sqrt(T/μ) where T = tension and μ = linear density. Substituting our given values into the formula, we get v = sqrt(40N / 4g/m) = 100 m/s.

The wavelength, λ, can be calculated using the formula λ = v/f, where f = frequency. Substituting our given values into the formula gives us λ = 100 m/s / 50 Hz = 2 m.

2. The amplitude of the wave is defined as the maximum displacement of the wave. In this case, the maximum displacement is given as 2.5 mm or 0.0025 m. Thus, the amplitude of the wave, A, is 0.0025 m.

In this situation, the actuator is at the center of it's height, moving downwards at t=0. Therefore, the wave is a cosine function and its phase constant φ is 0.

3. Now, let's determine the displacement equation D(x,t) of the travelling wave. The general form is as follows:

D(x,t) = A*cos(2πft - 2πx/λ + φ)

where D(x,t) = displacement as a function of x (position) and t (time), A = amplitude, f = frequency, λ = wavelength, and φ = phase constant.

Using the given values, this becomes:

D(x,t) = 0.0025*cos(314.159 t - 3.14159 x)

4. Lastly, we are asked to determine the string's displacement at x=0.50 m and t=10 ms. We can do this by substituting these values into our displacement equation:

D(0.50, 10e-3) = 0.0025*cos(314.159*10e-3 - 3.14159*0.50) ≈ 1.53e-19 m. Due to the wave's oscillation, it's extremely small - nearly zero - at this particular time and position.

User Jimj
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