Question

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.

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

With what speed would a pulse travel down the chain?

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

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

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

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.

What is the maximum speed of the 30th bead?

Answer 1

Stretched length of the chain, L = 1.50 m

mass of each bead, m = 1.00 g

number of beads, N = 101

mass of the chain, M = N*m = 101*1.0 g = 101 g

Spring constant of the strand, K = 28.8 N/m

1. linear mass density of the chain,

2. Extension produced in the chain,

Tension in the chain,

In significant figures, T = 14.3 N

3. Speed of the wave,

In significant figures, V = 14.6 m/s

4. It is the fourth mode of vibration, its frequency is:

5. As the string is vibrating with four antinodes, so there are four loops formed in the string with fixed ends, so it vibrates in four segments. Length of each segment is:

So the beads which are at rest are at locations:

x = 0, 375mm, 750mm, 1125mm and 1500 mm are at rest

Distance between the centres of the adjacent beads is

So the numbers of the beads are

6. Maximum velocity

Where A is amplitude

Therefore amplitude of 13th bead

(in significat figures)

7. Position of nth bead is given by the relation

Where d = 15 mm = 0.015m as explained earlier in part 5.

For n = 1

For n=2

For n =3

For n=4

8. location of 30th bead is

Amplitude of the wave at distance x is given by

Where Ao = Amplitude at the location of antinode.

= wavelength = 2*length of one loop = 2*375 mm=750 mm

Max speed

Comparing 30th and 13th bead

So maximum speed of 30th bead is 3.64 m/s