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A sphere of mass M 5 1.00 kg is supported by a string that passes over a pulley at the end of a horizontal rod of length L 5 0.300 m (Fig. P17.15). The string makes an angle u 5 35.08 with the rod. The fundamental frequency of standing waves in the portion of the string above the rod is f 5 60.0 Hz. Find the mass of the portion of the string above the rod.

Respuesta :

A fundamental frequency of 60 Hz, attached mass of 1 kg., rod length of

0.3 m gives the mass of the part of the string as 3.245 × 10⁻³ kg.

Response:

  • The mass of the string, m ≈ 3.245 × 10⁻³ kg

How can the mass of the part of the string be calculated?

The given parameters are;

Mass of the sphere, M = 1.00 kg

Length of the rod, L = 0.300 m

Angle the string makes with the rod, θ = 35°

The fundamental frequency of the standing wave in the portion of the

string above the rod, f₁ = 60.0 Hz

Required;

The mass of the portion of the string above the rod

Solution:

The fundamental frequency in a string is given as follows;

  • [tex]f_1 = \mathbf{\dfrac{\sqrt{\dfrac{T}{m/l} } }{2 \cdot l}}[/tex]

Where;

T = The tension in the string

m = The mass of the string

l = The length of the string

  • [tex]The \ tension \ in \ the \ string, \ T= \mathbf{\dfrac{M \cdot g}{sin(\theta)}}[/tex]

Which gives;

[tex]T= \mathbf{\dfrac{1 \times 9.81}{sin(35)}} \approx 17.1[/tex]

The tension in the string, T ≈ 17.1 N

  • [tex]The \ length \ of \ the \ string, \ l=\mathbf{\dfrac{L}{cos(\theta)}}[/tex]

Therefore;

[tex]l=\dfrac{0.300}{cos \left(35^{\circ} \right)} \approx 0.366[/tex]

Therefore;

[tex]60.0 = \mathbf{\dfrac{\sqrt{\dfrac{17.1}{m/0.366} } }{2 \times 0.366}}[/tex]

Which gives;

[tex]m = \dfrac{17.1}{(60.0 \times 2 \times 0.366)^2} \times 0.366 \approx 3.245 \times 10^{-3}[/tex]

  • The mass of the string, m ≈ 3.245 × 10⁻³ kg

Learn more about the fundamental frequency of an material here:

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