Question

A ski gondola is connected to the top of a hill by a steel cable oflength 620 m and diameter 1.5 cm. As thegondola comes to the end of its run, it bumps into the terminal andsends a wave pulse along the cable. It is observed that it took14 s for the pulse to travel the length ofthe cable and then return.
(a) What is the speed of the pulse?
m/s
(b) What is the tension in the cable?
N

Answer 1

Answer:

Explanation:

Given

Length of cable $L=620 m$

Diameter of cable $d=1.5 cm$

time taken to return to original position $T=14 s$

time taken to cover distance L

$t=\frac{T}{2}=7 s$

velocity

$v=\frac{L}{t}=\frac{620}{7}=88.57 m/s$

(b)Relation between velocity of wave Tension is

$v=\sqrt{\frac{T}{\mu }} , where \mu =$mass per unit Length

$T=v^2\cdot \mu$

$T=(88.57)^2\cdot \frac{m}{L}$

$T=(88.57)^2\cdot \frac{\rho AL}{L}$

where $\rho =density\ of\ steel =7850 kg/m^3$

$A=area\ of\ cross-section=\frac{\pi }{4}d^2=1.76\times 10^{-4} cm^2$

$T=(88.57)^2\cdot 7850\times 1.76\times 10^{-4}$

$T=10,883 N$

Answer 2

(a) The speed of the pulse in the steel cable is 88.57 m/s.

(b)  The tension in the steel cable is 10,858.1 N.

Time of motion of the wave

The time taken for the wave to travel a single wavelength is calculated as follows;

t = 0.5 x 14 s

t = 7 s

Speed of the wave

The speed of the wave is calculated as follows;

v = L/t

v = 620/7

v = 88.57 m/s

Tension in the cable

The tension in the table is calculated as follows;

$v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu$

where;

• μ is mass per unit length = m/L

μ = m/L

μ = ρV/L

μ = (ρAL)/L

μ = ρA

Where;

• ρ is density of steel = 7820 kg/m³
• A is area of the steel wire

A = πd²/4

A = π(0.015)²/4

A = 1.77 x 10⁻⁴ m²

The tension in the cable is calculated as follows;

T = v²μ = v²ρA

T = (88.57)² x 7820 x 1.77 x 10⁻⁴

T = 10,858.1 N

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