A small block of mass m is kept on a bigger block of mass M...

A small block of mass $m$ is kept on a bigger block of mass $M$ which is attached to a vertical spring of spring constant $k$ as shown in the figure. The system oscillates vertically. (a) Find the resultant force on the smaller block when it is displaced through a distance $x$ above its equilibrium position. (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude ? (c) What can be the maximum amplitude with which the two blocks may oscillate together?

Key Concepts

Simple Harmonic Motion

This concept involves the oscillatory motion of a mass attached to a spring, where the restoring force is directly proportional to the displacement from equilibrium. The system’s behavior is characterized by sinusoidal motion with a specific frequency and period determined by the spring constant and the mass involved. Understanding this allows for the analysis of how forces vary over time and displacement in an oscillating system.

Equilibrium in Spring?Mass Systems

In a spring-mass system, the equilibrium position is where the force due to the spring exactly balances other forces such as gravity. At this point, the net force on the mass is zero. The analysis of forces when the mass is displaced from equilibrium involves determining how both gravitational and spring forces work together to restore equilibrium, which is fundamental in understanding oscillatory behavior.

Normal Force in Accelerated Systems

When an object rests on a moving or accelerating surface, the normal force is not simply equal to the object's weight. Instead, it must account for additional accelerations imparted by the system’s motion. This concept is crucial for analyzing situations where the effective weight of an object is altered by the upward or downward acceleration of the supporting surface, leading to variations in the contact force between the two bodies.

Maximum Amplitude for Coupled Oscillators

In systems where two bodies are in contact and oscillate together, there exists a maximum amplitude beyond which the contact may be lost. This condition is determined by ensuring that the net acceleration acting on the object does not exceed the gravitational acceleration in a way that would cause separation. Recognizing the limits imposed by these dynamics is key to determining the maximum oscillation amplitude that the system can sustain while remaining coupled.

Transcript

00:02 Hello and welcome to this video solution of numerate.

00:05 Here we are given that a small block of mass small m is kept on a bigger block of mass capital m which is attached to a vertical spring of constant k.

00:14 So here we have a spring and over which we have a bigger block of mass capital m and above which we have the small mass m right and the system oscillates vertically.

00:36 So now we have to calculate the resultant force on the the smaller block when it is displaced to a distance of x above this equilibrium point now the whole system is moved upwards so let's say the bigger block moves here right and the distance between the extension is x now you have to calculate the forces on the uh result and force on this smaller block right now considering let me draw the smaller block block separately now here the forces acting will be the weight m g acting downwards right the reaction force are or the normal force in between these two blocks will act up up first and at the highest point it will move the overall acceleration will be a or the net force will be towards down right this we can say from this we can say that the net force is given by this force if which is equal to mass times the acceleration of the block right now acceleration of the block can we calculated as m omega square x and now you have to calculate what is the angular frequency of the system which is equal to root over of k by m right k by total mass here we will have m plus m this gives us equal to m k x over capital n plus small m so this is the first answer that is the overall the resulting force on the smaller mass next we have part b here you have to calculate normal force on this smaller block at this position okay so here we have the normal force so we can write the net force f equal to the mass m yes sorry the weight minus the normal reaction this gives us normal reaction equal to m g minus if now we plug in the value of f we have let's say we take m common from here so here g minus k x over capital m plus m so this is the normal force on this smaller block next we have a separate part c here you have to calculate the maximum amplitude with which the two blocks will oscillate together.

03:55 Here we have to calculate, we will write the net expression...

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