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A rope has a mass of 1.8 kg and a length of 11.1 m. It is fixed at both ends. If the frequency of the first harmonic on this rope is 1.09 Hz, what is the tension of this rope in N? Enter a number with one digit behind the decimal point.

Respuesta :

Answer:

Tension, T = 95 N

Explanation:

Given that,

Mass of the rope, m = 1.8 kg

length of the rope, L = 11.1 m

Frequency of the first harmonic, f = 1.09 Hz

The fundamental frequency in case of string is given by :

[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}[/tex]

[tex]\mu=\dfrac{m}{L}=\dfrac{1.8}{11.1}=0.162\ kg/m[/tex]

[tex]T=4L^2f^2\mu[/tex]

[tex]T=4\times (11.1)^2\times (1.09)^2\times 0.162[/tex]      `

T = 94.85 N

or

T = 95 N

So, the tension of this rope is 95 N. Hence, this is the required solution.

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