A 1.005 m chain consists of small spherical beads, each with a mass of 1.00 g and a diameter of 5.00 mm, threaded on an elastic strand with negligible mass such that adjacent beads are separated by a center-to-center distance of 10.0 mm. There are beads at each end of the chain. The strand has a spring constant of 28.8 N/m. The chain is stretched horizontally on a frictionless tabletop to a length of 1.50 m, and the beads at both ends are fixed in place.

1. What is the linear mass density of the chain? What is the tension in the chain?

2. With what speed would a pulse travel down the chain?

3. The chain is set vibrating and exhibits a standing-wave pattern with four antinodes. What is the frequency of this motion?

4. If the beads are numbered sequentially from 1 to 101, what are the numbers of the five beads that remain motionless?

5. The 13th bead has a maximum speed of 7.54 m/s. What is the amplitude of that bead's motion?

6. If x0=0 corresponds to the center of the 1st bead and x101=1.50m corresponds to the center of the 101st bead, what is the position xn of the nth bead? Calculate the first four coordinates.

7. What is the maximum speed of the 30th bead?

1) μ = 1.33 10⁻³ kg / m , F = - 14,256 ,  2) v= 103.53 m/s, 3)  f = 138.04 Hz , 4)  1, 25, 50, 76, 101   , 5) A = 0.00869 m , 6)  # _position = (# _account-1) (1.5m / 100 accounts)

Explanation:

1) Linear density is the mass per unit length

     μ = m / L

     μ = 2 1 10⁻³ / 1,5

     μ = 1.33 10⁻³ kg / m

this is the density when the chain is stretched, which is when the pulse occurs

we can find the tension with

     F = - k (x₁-x₀)

where k is the spring constant

     F = - 28.8 (1.5 -1.005)

     F = - 14.256 N

the negative sign indicates that the force is restorative

2) the pulse speed is

      v = √ T /μ

      v = √ 14,256 / 1,33 10⁻³

      v = 103.53 m / s

3) If standing waves are formed with fixed points at the ends and 4 antinodes, the wavelength is

          2 λ = L

            λ = L / 2

wave speed is related to frequency and wavelength

           v = λ f

            f = v / λ

            f = v 2 / L

            f = 103.53 2 / 1.5

            f = 138.04 Hz

4) The marbles are numbered, the marbles that remain motionless are

   the first (1) and the last (101)

Let's look for the distance to each node, for this we must observe that in each wavelength there is a node at the beginning, one in the center and one at the end, therefore the nodes are in

         #_node = m λ / 2 = m L / 4

        #_node     position (m)

         1                  1.5 / 4 = 0.375

         2             2 1.5 / 4 = 0.75

         3             3 1.5 / 4 = 1,125

Since there are 101 marbles in the initial length, this number does not change with increasing length, so there is 101 marble in 1.5 m. Let's find with a direct proportion rule the number of marbles at these points with nodes

        #_canica = 0.375 m (101 marble / 1.5 m) 0.375 67.33

        # _canica = 25

        #_canica = 0.75 67.33

        #_canica = 50

        # _canica = 1,125 67.33

        #canica = 75.7 = 76

in short the number of the fixed marbles is

      1, 25, 50, 76, 101 canic

5) The movement of the account is oscillatory at this point, which is why it is described by

          y = A cos wt

          = -A w sin wt

the speed is maximum for when the breast is worth ±1

          v_{y} = Aw

           A = v_{y} / w

angular velocity related to frequency

         w = 2π f

          A = v_{y} / 2πf

          A = 7.54 / (2π 138.04)

          A = 0.00869 m

6) for the position of each account we can use a direct proportion rule

      in total there are 100 accounts distributed in the 1.50 m distance, the #_account is in the # _position. Note that it starts to be numbered 1, so this number must be subtracted from the index of the amount

       # _position = (# _account-1) (1.5m / 100 accounts)

#_canic position(m)

   1          0

   2         0.015

   3         0.045

   4         0.06

7) the wave has a constant velocity, but every wave is oscillated perpendicular to this velocity, with an oscillatory movement described by the expression

         y = Acos wt

the maximum speed is

          = -Aw sin wt

speed is maximum when the sine is ±1

         v_{y} = A w

to calculate the amplitude of the count we use that for a standing wave

         y = 2Asin kx

          y / A = 2 sin (2π /λ x)

the wavelength is

 λ = 0.75 m

the position is

x (30) = 29 1.5 / 100 = 0.435  m

          y (30) A = 2 sin (2pi 0.435 / 0.75)

          y (30) / A = 0.96 m

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .

probably not if net force is zero