wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A mass m is suspended from a spring of spring constant K and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is


A
2πmK
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
πmK+π mK2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
π m3K2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
πmK+πm2K
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D πmK+πm2K
The mass m completes first half of oscillation (upward) under the effect of upper spring alone.
Since, T=2πmK, time taken for one half of oscillation t1=T2=πmK
When the mass goes downwards, it is under the combined action of both the springs.
Both springs are in parallel combination.
So, Keq=K1+K2=2K
Time taken to complete second half of oscillation
t2=T2=2πm2K2=πm2K
Therefore, total time for 1 complete oscillation
=t1+t2=πmK+πm2K

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