Two blocks of masses m and M,(M>m), are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released, then ( μ= coefficient of friction between the two blocks)
A. The time period of small oscillation of the two blocks is T=2 π √((m+M)/k)
B. The acceleration of the blocks is a=-(k x)/(M+m) ( x= displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is (m μ|x|)/(M+m)
D. The maximum amplitude of the upper block, if it does not slip, is (μ(M+m) g)/k
E. Maximum frictional force can be μ(M+m) g.
Choose the correct answer from the options given below :

Solution

Physics sir · 2 steps

No slipping means one body of mass M+m; friction only has to give the top block m times a.

1. Move as one. If the upper block does not slip, both blocks share the same displacement x and acceleration. Only the spring acts horizontally on the pair, with force -kx, so the total mass M+m oscillates on spring constant k.

   a = -kx/(M+m), T = 2π√((M+m)/k)

   Statement	Check	Verdict
   A	Mass M+m on spring k gives T = 2π√((M+m)/k)	correct
   B	(M+m)a = -kx, so a = -kx/(M+m)	correct
   C	Friction on m is m|a| = mk|x|/(M+m). It has k, not μ	wrong
   D	Largest friction needed is at |x| = A: mkA/(M+m) ≤ μ mg, so Aₘₐₓ = μ(M+m)g/k	correct
   E	Normal force on m is mg, so friction can be at most μ mg	wrong

2. So. Statements A, B and D are true, which is option (c).

The idea to keep. For stacked blocks on a spring, use the total mass to get ω, then find friction from f = ma on the top block alone. It is largest at the extremes and can never exceed μ mg.

Answer from the NTA final answer key. Solution by PrepWiser.