Answer :
Answer:
= 1.37 Hz
Explanation:
F = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]
Given that
Frequency of oscillation of the empty diving board when executing Simple harmonic motion = 4.00Hz
Mass of the diving board ,m= 10kg
Mass of the person on the diving board, M = 75kg
Let k be the stiffness constant of the diving board
Let F₂ be the frequency of the board with the person on it
Frequency of the diving board F₁ = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m} }[/tex]
Frequency of the diving board with the person on it
F₂ = [tex]\frac{1}{2\pi } \sqrt{\frac{k}{m + M} }[/tex]
Squaring both the equations and dividing the second equation with first equation
F₂² / F₁² = m / (m+M)
the frequency of the diving board with the person
F₂ = [tex]\sqrt{F^2\frac{m}{m + M} }[/tex]
= [tex]\sqrt{4^2\frac{10}{10 + 75} }[/tex]
= 1.37 Hz
the frequency of the simple harmonic motion when the diver is on the board is = 1.37 Hz
The frequency of the simple harmonic motion of the diver on the board is 1.372 Hz.
The given parameters;
- frequency of the board, F₀ = 4 Hz
- effective mass of the board, m₀ = 10 kg
- mass of the person on the board, m₁ = 75 kg
The relationship between frequency of oscillation and mass of the oscillating object is given as;
[tex]F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F_1\sqrt{m_1} = F_2\sqrt{m_2} \\\\F_2 = F_1\sqrt{\frac{m_1}{m_2} } \\\\F_2 = 4 \times \sqrt{\frac{10}{10 + 75} } \\\\F_2 = 1.372 \ Hz[/tex]
Thus, the frequency of the simple harmonic motion of the diver on the board is 1.372 Hz.
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