PLEASE HELP WITH THESE!!!! WILL GIVE RATE FOR GOOD WORK The displacement of a st

Question

PLEASE HELP WITH THESE!!!! WILL GIVE RATE FOR GOOD WORK

The displacement of a standing wave on a string is given by D=2.8sin(0.59x)cos(37t), where x and D are in centimeters and t is in seconds.

1)Give the speed of each of the component waves. v1,v2 in cm/s

2)Find the speed of a particle of the string at x=3.00cm when t=2.8s

3)A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 3.1 s later. How deep is the ocean (ANSWER IS NOT 2418m)

4)A 5.8 kg ball hangs from a steel wire 2.00 mm in diameter and 7.00 m long. What would be the speed of a wave in the steel wire?

Explanation / Answer

1)

Velocities, v1=v2 = omega/k = 37/0.59 = 62.71 cm/s

2)

Speed of particle at an instant = v = dD/dt = - 103.6 sin(0.59x)sin(37t) =

v at x=3.00cm , t=2.8s is v = -103.6 sin(0.59*3) sin (37*2.8) = 7.36 cm/s

3)

speed of sound v = sqrt(K/)

K - bulk modulus of water = 2.34×10^9 Pa
- density of seawater = 1,025 kilogram/m^3

v = 1510 m/s

d = (1510)(1.55) = 2341 meters (we use only 1.55seconds because the sound had to travel down to the ocean floor and then return)

4)

The wave speed in a wire = sqrt(T/mu) where T is tension and mu = the linear distribution of mass

Now T = weight hanging = m*g

mu = density of steel*volume of the wire/length of the wire.

This reduces to density * cross section area of the wire

So density of steel = 8050 kg/m3

Area = pi*(1x10^-3m)^2

so mu = 8050 kg/m3*(pi*(1x10^-3m)^2) = 0.0252 kg/m
Therefore v = sqrt(5.8*9.8/0.252) =47.42 m/s

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