wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4.5 Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
5 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
5.52 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon