Question

A 1.80-m string of weight 0.0126 N 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 equation y(x,t)=(8.50mm)cos(172rad⋅m−1x−2730rad⋅s−1t) Assume that the tension of the string is constant and equal to W.

Answer 1

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

                            y (x, t) = A*cos(k*x -w*t)  

Where k = stiffness and w = angular frequency

Hence,

                           k = 172 and w = 2730

- Calculate wave speed V:

                            V = w / k = 2730 / 172 = 13.78 m/s

- Tension in the string T:

                            T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

                           T = W = Y*V^2 = (w/L*g)*V^2

                            W = (0.0126 / 1.8*9.81)*(13.78)^2

                            W = 0.135 N