A mass of 5kg is suspended from a spring of stiffness constant 4000N/m. The system is fitted with a damper with a damping coefficient of 0.2. The mass is pulled down by 50mm and then released. Calculate the damped frequency.
A
4Hz
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B
4.5Hz
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C
5Hz
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D
5.52Hz
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Solution
The correct option is B4.5Hz The angular frequency of free oscillations is given by ω0=√km
From the data given in the question, we get ω0=√40005=28.28rad/s
Let ω′ be the angular frequency of damped oscillations.
Then, ω′=√ω20−γ2
Given, γ=0.2 ∴ω′=√(28.28)2−(0.2)2=28.27rad/s ⇒f′=ω′2π=28.272×3.14≈4.5Hz
Thus, option (b) is the correct answer.