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Sagot :
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:
https://brainly.com/question/3776467
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