A transverse pulse moves along a stretched cord of length 7.14 m having a mass of 0.120 kg. If the tension in the cord is 10.2 N, find the following.

a)wave speed
b)time to travel length of cord

Mass per unit length m₀ = m/L

Velocity in the stretched string is v = sqrt(T/m₀).
t=L/v

To find the wave speed (v) of the pulse on the cord, we can use the formula:

v = sqrt(T/u)

where T is the tension in the cord and u is the linear mass density.

To find the linear mass density (u), we divide the mass (m) of the cord by its length (L):

u = m/L

Let's plug in the given values:

m = 0.120 kg
L = 7.14 m
T = 10.2 N

u = 0.120 kg / 7.14 m = 0.0168 kg/m

Now we can find the wave speed:

v = sqrt(10.2 N / 0.0168 kg/m)

v = sqrt(607.142857 N/kg)

v ≈ 24.65 m/s

So the wave speed is approximately 24.65 m/s.

To find the time (t) it takes for the pulse to travel the length of the cord, we can use the formula:

t = L/v

Let's plug in the values:

L = 7.14 m
v ≈ 24.65 m/s

t = 7.14 m / 24.65 m/s

t ≈ 0.29 s

So it takes approximately 0.29 seconds for the pulse to travel the length of the cord.

To find the wave speed, we need to use the formula:

Wave speed (v) = sqrt(Tension (T) / linear density (μ))

a) To find the wave speed, we need to determine the linear density (μ) of the cord. The linear density is given by the mass per unit length of the cord:

Linear density (μ) = Mass (m) / Length (L)

Substituting the given values:
Mass (m) = 0.120 kg
Length (L) = 7.14 m

Linear density (μ) = 0.120 kg / 7.14 m

Now, we can calculate the linear density (μ) and substitute it into the wave speed formula:

Wave speed (v) = sqrt(10.2 N / (0.120 kg / 7.14 m))

b) To find the time taken to travel the length of the cord, we can use the formula:

Time (t) = Length (L) / Wave speed (v)

Substituting the given values:
Length (L) = 7.14 m
Wave speed (v) = calculated value from part a)

Time (t) = 7.14 m / calculated wave speed (v)

By plugging in the values into the formula, you can find the wave speed and time to travel the length of the cord.