A1.80-m string of weight 0.0123n is tied to the ceiling at its upper end, and the lower end supports a weight w. neglect the very small variation in tension along the length of the string that is produced by the weight of the string. when you pluck the string slightly, the waves traveling up the string obey the equationy(x, t)=(8.50mm)cos(172rad? m^(? 1)x? 2730rad? s^(? 1)t)assume that the tension of the string is constant and equal to w. a) how much time does it take a pulse to travel the full length of the string? b) what is the weight w? c) how many wavelengths are on the string at any instant of time? d) what is the equation for waves traveling down the string? a) y(x, t)=(8.50mm)cos(172rad? m? 1x? 2730rad? s? 1t)b) y(x, t)=(8.50mm)cos(172rad? m? 1x+2730rad? s? 1t)c) y(x, t)=(10.5mm)cos(172rad? m? 1x+2730rad? s? 1t)d) y(x, t)=(10.5mm)cos(172rad? m? 1x? 2730rad? s? 1t)

a)    t = 0.113 s , b)     W = 0.1756 N , c)   # λ = 49

Explanation:

a) Let's use the relationship

         v = λ f

Of the equation

         y = 8.55 10-3 cos (172 x + 2730 t)

When comparing this with the general equation

         y = A cos (kx - wt)

The wave number k = 172

       k = 2π /λ

       λ = 2π / 172

       λ = 0.03653 m

The angular velocity w = 2730

     w = 2π f

      f = w / 2π

     f = 2730 / 2π

     f = 434.49 Hz

The speed of the wave is

      v = 0.03653    434.49

      v = 15.87 m / s

The speed the wave on a string is constant, so

      v = d / t

      t = d / v

      t = 1.80 /15.87

      t = 0.113 s

b) The weight applied to the rope

      v = √ T /μ

The density

     μ = m / l

     μ = (0.0123 / 9.8) /1.80

     μ = 6.97 10-4 kg / m

The tension equal to the applied weight

    T = v² μ

    W = T = 15.87²   6.97 10⁻⁴

    W = 0.1756 N

c) let's use a rule of proportions

    # λ = 1.8 /0.03653

      # λ = 49

d.7.5cm is your answer

d. air pressure decreases as altitude increases.