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

A block A of mass m1 and block B of mass m2 are connected together by a spring and placed on a bearing surface as shown. The block A performs free vertical harmonic oscillations with amplitude Amax and frequency ω. Assuming that block B doesn't leave the surface and neglecting the mass of the spring, find the maximum and minimum value of force (Rmax & Rmin) that the system exerts on the bearing surface.


A
Rmax=(m1+m2)g+m1ω2Amax; Rmin=(m1+m2)gm1ω2Amax
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Rmax=(m1+m2)g+m2ω2Amax; Rmin=(m1+m2)gm2ω2Amax
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Rmax=Rmin=(m1+m2)g+m1ω2Amax
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Rmax=Rmin=(m1+m2)gm1ω2Amax
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A Rmax=(m1+m2)g+m1ω2Amax; Rmin=(m1+m2)gm1ω2Amax
As A oscillates up and down, the normal reaction between B and the surface is maximum when A is at the lowest position and it is minimum when A is at the topmost position.


Case I: Lowest position of A
Let T1 be the force in the compressed spring.
As B is at rest, we can say that,
T1+m2g=Rmax ........(1)
As acceleration of A is amax=ω2Amax towards mean position () (where Amax is amplitude),
T1m1g=m1ω2Amax .......(2)
From (1) and (2),
Rmax=(m1+m2)g+m1ω2Amax ......(3)

Case II: Topmost position of A


Let T2 be the force in the elongated spring.
As B is at rest,
T2+Rmin=m2g
Acceleration of A is amax=ω2Amax towards mean position (), where Amax is amplitude.
T2+m1g=m1ω2Amax
Combining, we get Rmin=(m1+m2)gm1ω2Amax ......(4)
From (3) and (4), we can say that option (a) is the correct answer.

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon