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Question

A mass of 5 kg is suspended from a spring of stiffness constant 4000 N/m. The system is fitted with a damper with a damping coefficient of 0.2. The mass is pulled down by 50 mm and then released. Calculate the damped frequency.

A
4 Hz
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B
4.5 Hz
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C
5 Hz
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D
5.52 Hz
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Solution

The correct option is B 4.5 Hz
The angular frequency of free oscillations is given by
ω0=km
From the data given in the question, we get
ω0=40005=28.28 rad/s
Let ω be the angular frequency of damped oscillations.
Then,
ω=ω20γ2
Given, γ=0.2
ω=(28.28)2(0.2)2=28.27 rad/s
f=ω2π=28.272×3.144.5 Hz
Thus, option (b) is the correct answer.

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