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 HzThe 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′=√ω20−γ2 Given, γ=0.2 ∴ ω′=√(28.28)2−(0.2)2=28.27 rad/s ⇒f′=ω′2π=28.272×3.14≈4.5 Hz Thus, option (b) is the correct answer.

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program