Irodov Problem 4.23

The figure depicts the manner in which the mass m undergoes longitudinal oscillations in the spring. As the mass m is moves to the right, the right part of the spring is compressed while the left part of the spring expands. Thus, the left side of the spring pulls the mass to the left and the right part of the spring pushes the mass to the left.

The spring is an elastic material and follows the Young's law. In other words, if the Young's modulus of the springs material is , its area of cross section is A, then for any section of the spring we have,



Let us first consider the entire spring. Lets its length be l0. Then, the restoring force resulting from a compression of is given by,



Hooke's law for springs says that,



Comparing (2) and (3) we have,



Now let us consider what happens when the mass m is pushed a distance x to the right.

The left part of the spring is expanded by an amount x. Since the length of the left part of the spring is , the restoring force for this part of the spring can be computed using (2) as,



The right part of the spring is compressed by an amount x. Since the length of the right part of the spring is , the restoring force for this part of the spring can be computed using (2) as,



Thus the net restoring force acting on the particle will be,



From (7) it is clear that, the particle is undergoing simple harmonic motion with,

Irodov Problem 4.22

There are two forces acting on the hydrometer i) the force due to gravity mg pulling it down into the liquid and ii) the buoyant force B acting upwards from the liquid trying to push it out of the liquid. The buoyant force acting on the hydrometer is equal to the weight of the water displaced by the submerged part of the hydrometer.


At equilibrium, the hydrometer is at rest and so B = mg. When the hydrometer is pushed down into the liquid by a distance x beyond the equilibrium point, the hydrometer displaces and additional amount of water and so the Buoyant force increases to B' > B. Thus, the hydrometer is pushed out. Specifically, the additional volume of liquid displaced when the hydrometer is pushed down by a value x is and so the weight of the additional liquid displaced is . Thus,



Similarly when the hydrometer is pulled a distance x out of the liquid, the volume of liquid displaced by the hydrometer decreases and so the buoyant force B'' becomes less than B. Thus, the wieght of the hydrometer dominates and pulls the hydrometer back into the liquid.

The net force acting on the hydrometer when its pushed x distance below the equilibrium point is given by,



We already know that the equation of motion for a body undergoing simple harmonic motions is given by,



where is the angular frequency. Hence, comparing this with (2) we have,

Irodov Problem 4.21

When the lift accelerates upwards at a rate w, it is equivalent to the situation where acceleration due to gravity were g+w instead of g. The time period of the pendulum in the upward accelerating lift is given by,



Since the time period of the pendulum under normal conditions should be,



during the time when the elevator is accelerating upwards, the clock runs faster (since the time period as given by (1) is smaller) by a factor .

When the list accelerates downwards at a rate w, it is equivalent to the situation where the acceleration due to gravity is g-w. The time period of the pendulum when the list is accelerating downwards is given by,



Thus, during the time when the elevator is accelerating downwards, the clock runs slower (since the time period as given by (2) is longer) by a factor .


The time required by the list to travel a height h starting from rest while accelerating at a rate w is given by,



The time shown by the clock at the end of this interval is going to be,



Now let us assume that the elevator travels downwards for a time duration before the clock shows the correct time. The corresponding time duration as seen in the downward accelerating clock will be given by,



For the time to be correct, at the end of then,



The total time the elevator traveled is then given by,

Irodov Problem 4.20

We know that for small oscillation angles the angle of the pendulum as a function of time is given as,



The time taken for the bob to reach the wall is given by,



Since the collision between the wall and the pendulum is elastic, it reflects back with exactly the same velocity as it hits the wall. Thus, the total time period of the pendulum bob is double the time taken to reach the wall and is given by,

Irodov Problem 4.19

There are three forces acting on the pendulum bob, i) the gravitational force mg, acting vertically downwards ii) the buoyant force acting on the bob vertically upwards and iii) the tension in the string T. Since the density of the fluid is times less denser, the buoyant force acting on the bob is .

The net force acting on the bob in the direction perpendicular to the thread is and hence the torque acting on the bob about the point of suspension is given by . Assuming that the pendulum bob is so small that it can be considered to be a point mass, its moment of inertia about the point of suspension is given by . Hence we have,







For small values of we have,





From two it is clear that the oscillations are simple harmonic with a period of

Irodov Problem 4.18

The component of the tension in the string (F) along the x-axis ( ) is responsible for providing the restoring force to move the particle towards the center. The net restoring force is thus given by due to each half of the string.

From elementary geometry we have,




Hence the equation of motion of the particle is given by,