Two blocks of masses m and M, (M > m)$(\mathbf{M} > \mathbf{m})$ , 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)$(\mu =\text{coefficient of friction between the two blocks})$The diagram displays a smaller block of mass m placed on top of a larger block of mass M, which is connected to a horizontal spring on a smooth table.
(A) The time period of small oscillation of the two blocks is T = 2π (m + M)k$\mathrm{T} = 2\pi \sqrt{\frac{(\mathrm{m} + \mathrm{M})}{\mathrm{k}}}$
(B) The acceleration of the blocks is a = (kx)/(M + m)$a = \frac{kx}{M + m}$ (x = displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is mk|x|M + m$\frac{\mathrm{m}k|\mathrm{x}|}{\mathrm{M} + \mathrm{m}}$
(D) The maximum amplitude of the upper block, if it does not slip, is μ(M + m)gk$\frac{\mu(\mathbf{M} + \mathbf{m})\mathbf{g}}{\mathbf{k}}$
(E) Maximum frictional force can be μ (M + m)g$\mu (\mathbf{M} + \mathbf{m})\mathbf{g}$.
Choose the correct answer from the options given below:
A.A, B, D Only
B.B, C, D Only
C.C, D, E Only
D.A, B, C Only
Solution & Explanation
Related Formula
For combined system performing simple harmonic motion without relative slipping:
T = 2π mtotalk$$T = 2\pi \sqrt{\frac{m_{\text{total}}}{k}}$$a = -ω² x = -(k)/(M+m) x$$a = -\omega^2 x = -\frac{k}{M+m} x$$
Core Logic
Let's analyze each statement:
Statement (A): Since both blocks perform SHM together, the combined mass is (M + m)$(M + m)$. The spring constant is k$k$. Thus, the time period of small oscillation is:
T = 2π √((M+m)/(k))$$T = 2\pi \sqrt{\frac{M+m}{k}}$$
This is correct. (A is True)
Statement (B): When the system is displaced by x$x$, the restoring spring force on the combined system is F = -kx$F = -kx$. The common acceleration of the combined mass is:
This matches the expression (taking magnitude). (B is True)
Statement (C): The upper block of mass m$m$ moves solely due to the static frictional force f$f$ acting on it. Thus:
f = m a = m ( (kx)/(M+m) ) = (mkx)/(M+m)$$f = m a = m \left( \frac{kx}{M+m} \right) = \frac{mkx}{M+m}$$
Statement (C) claims the frictional force is (mμ|x|)/(M+m)$\frac{m\mu|x|}{M+m}$, which is incorrect because friction is determined by acceleration, not by coefficient of friction μ$\mu$ during static grip. (C is False)
Statement (D): For no slipping to occur, the maximum frictional force required at peak amplitude A$A$ must be less than or equal to the limiting static friction fL = μ mg$f_L = \mu mg$:
fmax = (mkA)/(M+m) ≤ μ mg$$f_{\text{max}} = \frac{mkA}{M+m} \le \mu mg$$(kA)/(M+m) ≤ μ g A ≤ (μ(M+m)g)/(k)$$\frac{kA}{M+m} \le \mu g \implies A \le \frac{\mu(M+m)g}{k}$$
Thus, the maximum amplitude is (μ(M+m)g)/(k)$\frac{\mu(M+m)g}{k}$. (D is True)
Statement (E): The maximum static frictional force between the blocks is fL = μ mg$f_L = \mu mg$, not μ (M+m)g$\mu (M+m)g$. (E is False)
Step 1: Conclusion
Only statements A, B, and D are correct. Hence, the correct option is (1).
The diagram displays a smaller block of mass m placed on top of a larger block of mass M, which is connected to a horizontal spring on a smooth table.The diagram displays a smaller block of mass m placed on top of a larger block of mass M, which is connected to a horizontal spring on a smooth table.
Pattern Recognition
In stacked blocks with springs, always identify the force driving the non-spring-loaded block. Here, mass m$m$ is driven purely by friction, so f = m · a$f = m \cdot a$. Slipping begins when this required force exceeds flimit = μ m g$f_{\text{limit}} = \mu m g$. This simple boundary matches the derivation of maximum amplitude perfectly!
Chapter Mix
Class 11 Physics: Laws of Motion: Friction
Class 11 Physics: Oscillations: Simple Harmonic Motion
Keywords:#stacked block SHM friction#JEE Main 2025 Morning Q6#maximum amplitude no slipping#Oscillations and friction JEE#spring block system#stacked blocks#frictional oscillation
More Laws of Motion Previous-Year Questions
Q31jee_main_2026_21_jan_morningForce and Acceleration
A 4 kg mass moves under the influence of a forceF = (4t³ i - 3t j) N$\vec{\mathrm{F}} = (4\mathrm{t}^3\hat{\mathrm{i}} - 3\mathrm{t}\hat{\mathrm{j}})\text{ N}$ where t is the time in second. If mass starts from origin at t = 0$t = 0$, the velocity and position after t = 2s$t = 2\text{s}$ will be:
A.v = 3 i +(3)/(2) j, r = (6)/(5) i + j$\vec{\mathrm{v}} = 3\hat{\mathrm{i}} +\frac{3}{2}\hat{\mathrm{j}}, \quad \vec{\mathrm{r}} = \frac{6}{5}\hat{\mathrm{i}} +\hat{\mathrm{j}}$
B.v = 4 i -(3)/(2) j, r = (8)/(5) i - j$\vec{\mathrm{v}} = 4\hat{\mathrm{i}} -\frac{3}{2}\hat{\mathrm{j}}, \quad \vec{\mathrm{r}} = \frac{8}{5}\hat{\mathrm{i}} -\hat{\mathrm{j}}$
C.v = 4 i +(5)/(2) j, r = (8)/(5) i +2 j$\vec{\mathrm{v}} = 4\hat{\mathrm{i}} +\frac{5}{2}\hat{\mathrm{j}}, \quad \vec{\mathrm{r}} = \frac{8}{5}\hat{\mathrm{i}} +2\hat{\mathrm{j}}$
D.v = 4 i -(3)/(2) j, r = (6)/(5) i - j$\vec{\mathrm{v}} = 4\hat{\mathrm{i}} -\frac{3}{2}\hat{\mathrm{j}}, \quad \vec{\mathrm{r}} = \frac{6}{5}\hat{\mathrm{i}} -\hat{\mathrm{j}}$
Solution
Related Formula
a = Fm$$\vec{a} = \frac{\vec{F}}{m}$$v = ∫ a dt$$\vec{v} = \int \vec{a} \, dt$$r = ∫ v dt$$\vec{r} = \int \vec{v} \, dt$$
Core Logic
Given F = 4t³ i - 3t j$\vec{F} = 4t^3\hat{i} - 3t\hat{j}$ and mass m = 4 kg$m = 4\text{ kg}$.
Acceleration:
a = Fm = 4t³ i - 3t j4 = t³ i - (3)/(4)t j$$\vec{a} = \frac{\vec{F}}{m} = \frac{4t^3\hat{i} - 3t\hat{j}}{4} = t^3\hat{i} - \frac{3}{4}t\hat{j}$$
Step 1: Calculating Velocity
Velocity is the integral of acceleration with v(0) = 0$v(0) = 0$:
v = ∫ ( t³ i - (3)/(4)t j ) dt = (t⁴)/(4) i - (3)/(8)t² j$$\vec{v} = \int \left( t^3\hat{i} - \frac{3}{4}t\hat{j} \right) dt = \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j}$$
Position is the integral of velocity with r(0) = 0$\vec{r}(0) = \vec{0}$:
r = ∫ v dt = ∫ ( (t⁴)/(4) i - (3)/(8)t² j ) dt$$\vec{r} = \int \vec{v} \, dt = \int \left( \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j} \right) dt$$r = (t⁵)/(20) i - (3)/(24)t³ j = (t⁵)/(20) i - (t³)/(8) j$$\vec{r} = \frac{t^5}{20}\hat{i} - \frac{3}{24}t^3\hat{j} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$$
At t = 2 s$t = 2\text{ s}$:
r(2) = ((2)⁵)/(20) i - ((2)³)/(8) j = (32)/(20) i - (8)/(8) j = (8)/(5) i - j$$\vec{r}(2) = \frac{(2)^5}{20}\hat{i} - \frac{(2)^3}{8}\hat{j} = \frac{32}{20}\hat{i} - \frac{8}{8}\hat{j} = \frac{8}{5}\hat{i} - \hat{j}$$
Pattern Recognition
Recognize F → a → v → r$F \to a \to v \to r$ means consecutive integrations. Because the initial state is from rest at origin, we don't have to worry about constants of integration.
Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
Q43jee_main_2026_21_jan_eveningCircular Motion
A large drum having radius R is spinning around its axis with angular velocity ω$\omega$, as shown in figure. The minimum value of ω$\omega$ so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M is μ$\mu$, is :
Large drum of radius R spinning about its vertical axis with mass M stuck to the inner wall.
A.√((μ g)/(R))$\sqrt{\frac{\mu g}{R}}$
B.√((2g)/(μ R))$\sqrt{\frac{2g}{\mu R}}$
C.√((g)/(2μ R))$\sqrt{\frac{g}{2\mu R}}$
D.√((g)/(μ R))$\sqrt{\frac{g}{\mu R}}$
Solution
Related Formula
N = Mω² R$N = M\omega^2 R$
fmax = μ N$$f_{\text{max}} = \mu N$$
f = Mg$f = Mg$
Core Logic
Large drum of radius R spinning about its vertical axis with mass M stuck to the inner wall.
For the mass M$M$ to remain stuck to the inner wall and not slide down, the upward frictional force must balance the downward gravitational force.
f = Mg$f = Mg$
The normal force N$N$ providing the friction is generated by the centrifugal effect (or provides the centripetal acceleration):
N = Mω² R$N = M\omega^2 R$
The condition for no sliding is that required friction cannot exceed the maximum static friction:
Mg ≤ μ N$Mg \le \mu N$
In a "rotor" ride setup, gravity is opposed by friction, and the normal force is solely centripetal. Equating μ Fc$\mu F_c$ to mg$mg$ always yields the minimum spin limit.
Chapter Mix
Class 11 Physics: Laws of Motion
Q28jee_main_2026_23_january_eveningMotion on Inclined Plane
A block is sliding down on an inclined plane of slope θ$\theta$ and at an instant t = 0$t = 0$ this block is given an upward momentum so that it starts moving up on the inclined surface with velocity u$u$ . The distance (S) travelled by the block before its velocity become zero, is ____. (g = gravitational acceleration)
A.(u²)/(4g θ)$\frac{u^2}{4g\sin\theta}$
B.(2u²)/(g θ)$\frac{2u^2}{g\cos\theta}$
C.u²√(2)g θ$\frac{u^2}{\sqrt{2}g\cos\theta}$
D.(u²)/(2g θ)$\frac{u^2}{2g\sin\theta}$
Solution
Related Formula
v² = u² + 2as$v^2 = u^2 + 2as$
Core Logic
Motion on Inclined Plane diagram for Q28 - JEE Main 2026 Evening
Since the surface is frictionless (implied as no friction coefficient is given), the only force along the incline opposing the upward motion is the component of gravity, mg θ$mg \sin \theta$.
Acceleration a = -g θ$a = -g \sin \theta$.
Step 1: Apply Kinematic Equation
Using the third equation of motion:
V² = U² + 2as$V^2 = U^2 + 2as$
The spring of total length $\ell$ is cut into two parts with a length ratio of 1:3.
The lengths of the two pieces are:
₁ = (1)/(4)$\ell_1 = \frac{1}{4} \ell$ (smaller piece)
₂ = (3)/(4)$\ell_2 = \frac{3}{4} \ell$ (larger piece)
Step 1: Calculate Spring Constant of Smaller Piece
Since K = K' '$K \ell = K' \ell'$, for the smaller piece:
K = K' (( )/(4))$$K \ell = K' \left(\frac{\ell}{4}\right)$$
K' = 4K$K' = 4K$
Given K = 15 N/m$K = 15 \text{ N/m}$:
Spring constant is inversely proportional to spring length. Cutting a spring to 1/n$1/n$ of its length increases its stiffness by a factor of n$n$.
Chapter Mix
Class 11 Physics: Laws of Motion
Q49jee_main_2026_24_january_morningFriction
In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force F$\vec{\mathrm{F}}$ required to slide the block C with constant speed is ____ N. (Used g = 10 m/s²$g = 10 \text{ m/s}^{2}$)
Three stacked blocks A, B, C with an external pulling force F applied to C.
Three stacked blocks A, B, C with an external pulling force F applied to C.
Since C moves with constant speed, a=0$a=0$, meaning net force is zero.
For block A (4 kg$4\text{ kg}$) over B, sliding friction fA = μ (mA)g = 0.5 × 4 × 10 = 20 N$f_A = \mu (m_A)g = 0.5 \times 4 \times 10 = 20\text{ N}$.
For block B (6 kg$6\text{ kg}$) over C, the normal force includes block A. NB = (mA + mB)g = 100 N$N_B = (m_A + m_B)g = 100\text{ N}$.
Friction on B from C is fB = 0.5 × 100 = 50 N$f_B = 0.5 \times 100 = 50\text{ N}$.
For block C (8 kg$8\text{ kg}$) over ground, normal force is total weight NC = (mA + mB + mC)g = 180 N$N_C = (m_A + m_B + m_C)g = 180\text{ N}$.
Friction on C from ground is fC = 0.5 × 180 = 90 N$f_C = 0.5 \times 180 = 90\text{ N}$.
Step 1: Force Balance Equations
For block B to move relative to A and C (assuming standard string configuration where pulling C left causes B to be dragged right relative to C via pulley):
The tension T$T$ on B balances friction from A and C:
For block C moving left, the force F$F$ must overcome friction from ground, friction from B, and the tension T$T$ attached to it (depending on pulley setup. If pulley is on wall, string pulls C right):
Assuming standard setup where string connects B and C around wall pulley:
In multi-block pulley systems, pulling the bottom block accumulates all nested friction forces. The bottom block fights the ground, the block above it, AND the tension of the string anchoring the block above it.
Chapter Mix
Class 11 Physics: Laws of Motion
More Laws of Motion Questions — jee_main_2025_03_april_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
