Showing posts with label Time Period. Show all posts
Showing posts with label Time Period. Show all posts

Oscillations Problems with Solution Three

We are solving series of problems based on the concept of oscillation and simple harmonic motion. If the acceleration of a body is directly proportional to the displacement and opposite to it, then the motion is said to be simple harmonic motion. This can be done by many type of systems and if they are obeying the above mentioned condition, we can get the time period or frequency and it is constant all over the oscillatory motion. A loaded spring do oscillate when it is slightly disturbed. In practical way, any body won’t continue its oscillatory motion for ever and and due to air resistance, it slowly decreases and finally comes to the state of the rest. This kind of motion is called damped oscillatory motion and it that has to be continued, it shall have some external force support and that kind of motion is called forced oscillation.

Problem

We need to find the time period of pendulum of infinite length and the problem is as shown in the diagram below.


Solution

We need to write the equation for the torque as the product of force and perpendicular distance. Force is nothing but the weight of the body and distance is found as shown in the diagram below. We need to write further torque as the product of moment of inertia and angular acceleration basing on its definition. Thus we can equation for the acceleration and hence the time period as shown in the diagram below.


Problem

Frequency of a particle in SHM is given to us as shown below. We need to find the maximum speed that the particle can reach in the oscillatory motion.


Solution

The restoring force acting on the spring is nothing but the weight of the body and hence we can find the maximum possible displacement by equating them. By comparing that with the standard equation, we can find angular velocity and hence the maximum speed of the particle as shown in the diagram below.


Problem

A particle of known mass is attached to three springs as shown in the diagram below. If the particle is slightly disturbed, we need to know the time period of the system.

 
Solution

The resultant force acting on the mass using the vector laws of addition. We need to resolve the force into components and add as shown in the diagram below. By equating it to the restoring force, we can write the equation for the time period as shown in the diagram below.


Problem

Two blocks are kept one over the other and the lower surface is smooth and the connecting surface is rough as shown in the diagram below. The system is connected to a rigid support with a spring and the time period of the system is given to us in the problem. We need to find the mass of the upper block and the coefficient of friction so that there is no slipping between the two blocks.


Solution

We know the equation for the time period of a loaded spring and using that data, we can find the mass of the upper body as shown in the diagram below. Further equating the frictional force to the restoring force, we can solve the problem as shown in the diagram below.


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Oscillations Problems with Solution Two

We are solving series of problems based on  the concept of oscillations. It is also called vibratory or harmonic motion where there  is to and fro motion that gets repeated at regular intervals of time. The vertical projection of uniform circular motion is simple harmonic and we have derived equations for displacement, velocity and acceleration for the body in simple harmonic motion based on that. The time taken to complete one oscillation is called time period and the number of oscillations per one second is called frequency.

Problem

A body of known mass is connected with two springs and they in tern are connected to rigid support as shown in the diagram below. We need to find the effective time period of the system and the problem is as shown in the diagram below.


Solution

When ever the body is slightly disturbed, it starts oscillating and one spring expands and the other spring contracts by the same magnitude. As the force acting on both of them is same, the two springs behaves as if like they are connected in series and we can find the effective spring constant of the system and time period as shown in the diagram below.


Problem

A spring of spring constant K and length L is cut into two parts and the relation between the lengths of two parts and their ratio is given to us in the problem as shown in the diagram below. We need to measure the spring constant of one part of the spring.


Solution

Spring constant is the measure of nature of spring and it depends on the length of spring in the inverse proportional ratio. Taking that into consideration, we need to solve the problem as shown in the diagram below.


Problem

A piece of wood known dimensions is given to us and its density is also known to us as given in the problem below. It is floating in water with one surface vertical to the surface. It is pushed down and released and it starts oscillating. We need to measure the time period of the system.


Solution

When ever a force is applied on the body, there is buoyant force also and the system starts executing oscillatory motion with a certain restoring force. By comparing that with the standard equation, we can solve the problem as shown in the diagram below.


Problem

A cylindrical piston is used to close a cylinder with a certain gas and when it is slightly distributed, it starts oscillating in simple harmonic motion. We need to find the time period of the system and the problem is as shown in the diagram below.


Solution

By comparing it with the standard equation, we can solve the problem as shown in the diagram below. We need to identity the restoring force and the equation for the acceleration of the body in SHM. The problem is solved as shown in the diagram below. Here in the first case, we get the equation for the force.


We need to equate to the product of mass and acceleration and we further need to substitute the value of acceleration for a body in SHM so that we can compare it with the standard equation. Thus we can get the time period of the system as shown in the diagram below.


Problem

A spherical ball of known mass and radius is rolling with out slipping on a concave surface of known radius as shown in the diagram below. If the oscillations are small, we need to find the time period of the system.


Solution

Let us consider the spherical ball is at a particular position and we can find the angle basing on the definition as shown in the diagram below. As we know the equation for the acceleration of a body sliding on a inclined surface, by writing that equation, we can get that and find the time period as shown in the diagram below.



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Oscillations Problems with Solution One

We are solving series of problems based on the concept of oscillations. Oscillation is a kind of motion where the body oscillates about a fixed point called mean position and all oscillatory motions are periodic. It means oscillatory motion is repeated at regular intervals of time. We need to understand that all oscillatory motions are periodic but all periodic motions are not oscillatory. If oscillatory motion is also satisfying a condition like displacement is directly proportional to acceleration and acceleration is always directed to wards the mean position, we call that kind of oscillatory motion as simple harmonic motion. Simple pendulum is one example that executes simple harmonic motion when it is sightly disturbed from its mean position. We have derived equation for displacement, velocity and acceleration for a body in simple harmonic motion.

Problem

To a body in a simple harmonic motion, velocity is represented as shown in the equation below. We need to measure maximum acceleration that the body can get in the given conditions.



Solution

We have all ready derived equation for the velocity of the particle in simple harmonic motion. We need to get the given equation in the terms of the standard equation and the problem can be solved as shown in the diagram below.



Problem

A particle starts from mean position to a new position and it is as shown in the diagram below. Its amplitude and time period is given to us in the problem. We need to find the displacement where the velocity is half of the maximum velocity.


Solution

We know that the particle in simple harmonic motion has maximum velocity at the mean position. As per the given problem at a given instant, velocity of the particle is half of that maximum. Taking that into consideration and substituting the data in the standard format, we can solve the problem as shown in the diagram below.


Problem

Two different particles are in simple harmonic motion and their displacements are represented  as the given equations of the problem. We need to find the resultant amplitude of the combination. Problem is as shown below.


Solution

When we add to oscillatory motion, we need to get a oscillatory motion. The resultant amplitude can be found using the vector addition equation and the solution is as shown in the diagram below.


Problem

A simple harmonic oscillator starts from extreme position and covers a half the displacement in a given time. We need to measure the further time it is going to take to reach the mean position and the problem is as shown in the diagram below.


Solution

As the particle is here starting from the mean position, we need to know that it has some initial phase that is ninety degree. We know that the particle takes one forth of the time period to reach from extreme to mean position and to measure the remaining time to cover half amplitude to, we need to subtract from it as shown in the diagram below.


Problem

Number of springs are connected in series as shown in the problem to a a given mass and the system is allowed to oscillate. We need to measure the time period of oscillation of that system.


Solution

We know that when the springs are in series, the force acting on all of them is same and the extension in the spring is different and it depends on the nature of the spring. Using the common formula for the time period of the system and further simplify the problem as shown in the diagram below.



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Motion in One Dimension Problems with Solutions Two

We would like to solve the first problem basing on one dimensional body. The problem is about a body thrown initially with zero velocity. After two seconds another body is dropped with a certain velocity. We need to know the velocity of that body so that both the bodies reach the ground simultaneously.

Problem One


Solution

The first body is initially has zero velocity and using the second equation of motion, we can find the time taken by the body to reach the bottom of the bridge and strike the water. It is given that the second stone is thrown down with a initial velocity after two seconds. So the second body takes only three seconds to reach the ground. Both the stones reach the bottom of the bridge so that the distance travelled by both of them is same.

By substituting them in the same second equation of motion, we can find the initial velocity with which the second stone is thrown from the same place as shown in the diagram below.


Problem Two

This problem of one dimensional motion along Y axis is also of the similar  model of the previous problem. But in this problem, both the stones are falling freely so that have no initial velocity. The second stone is thrown few seconds later and we would like to know the distance of separation between the stones has to be a certain value after a specified time and we would like to find it out.


Solution

As per second equation of motion, we have relation between displacement, initial velocity, acceleration and time as shown in the diagram below. The second equation is different from the first as there is a time gap between both the stones.

It is given that after a certain time, we need to have a certain difference of separation. Thus by subtracting them we can solve the problem as shown in the diagram below.


Problem Three

This problem is also about two stones falling from a certain height and the second stone is having a falling distance certain separation 10 meter when compared with the first one.

It is given in the problem that both of them reach the bottom of the tower at the same time and the second stone is actually thrown one second later than the first one.


Solution

For a freely falling body, by using the second equation of motion, we can find the height travelled by the body  before it had reached the bottom. For the second stone also, we shall apply the same equation, but the time is one second less and height is 10 meter less.

By substituting the value of the height from the first equation in the second equation, we can solve the problem as shown in the diagram below.We can find both the time as well as height of the tower.


Problem Four

The problem is about the distance travelled by the body in the particular second. We have forth equation of motion in one dimensional motion. It is given that in the given second a certain distance of seven meter and in the previous second five meter has fallen. We would like to measure the velocity of the body with which it strikes the ground.


Solution

By solving them using forth equation of motion as shown in the diagram below after dividing both the cases, we can get the value of the n th second. Further we can find the final velocity of the body using the first equation of motion. The solution is shown in the diagram below.


Problem Five

It is given in the problem that one stone is dropped from a certain height 180 meter. It has no initial velocity. Another stone is dropped from a certain point below the first height and it means, it is travelling less distance than the first one. Both of them reached the ground at the same time but the second one got started initially late by two seconds.


Solution

By using the second equation of the motion, we can write two equations as shown in the diagram below. By solving both the equations, we can solve the problem as shown in the diagram below.


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Horizontal Projectile, Applications Problems with Solutions

A body projected horizontally from a certain height with an initial horizontal velocity can be called as a horizontal projectile. Its initial velocity along the vertical direction is zero and it possess only horizontal velocity at the beginning. As the time progresses, due to the impact of the gravity, it acquires the vertical component of velocity also. It can be shown that path taken by this body is parabola using the equations of motion of kinematics. We can write the equations for the displacement along x-axis and y-axis and further substituting the value of the time from the x-axis equation in the y-axis equation, it can be proved that its path is parabola.

At any instant of the path, velocity of the projectile is tangential to its path. It has both horizontal and vertical components and the angle made by the velocity vector with the horizontal can be calculated as shown below. We can also calculate the effective velocity of the projectile at that instant.


Application

Let us consider a body projected horizontally from the top of a tower. The line joining the point of projection and the striking point of ground makes an angle of 45° with the horizontal.  What is the displacement of the body?

Solution

As the angle made at the horizontally is 45 degree, according to trigonometry, its horizontal and vertical distances are equal. Being the displacement is a vector quantity with can calculate its resultant value using the parallelogram law of the vectors as shown below.


Application

Let us consider a ball is thrown horizontally from a staircase as shown with a initial velocity. We shall calculate how many steps it travels before it strikes the ground?

Solution

The ball is having only horizontal component of velocity and has no initial vertical component of velocity.

We can use equations of motion to find the displacement along x-axis and y-axis. There is no acceleration due to gravity impact on the x-axis and there is gravity acting along the y-axis. The total horizontal distance travelled by the body is equal to the multiplication of the number of the steps and the breadth of each step. Similarly the total vertical distance travelled by the body is equal to the multiplication of number of steps with the height of each step.

By substituting this values in the above equation is can solve the problem as shown below.


Application:

Two bodies are thrown horizontally with a two different initial velocities in mutually opposite directions from the same height. What is the time after which velocity is of the two bodies are perpendicular to each other ?

Solution:

Velocity is a physical quantity that is having both magnitude and direction and it has to be treated like a vector quantity. When two vectors are perpendicular to each other their scalar product becomes zero. Taking this into consideration and by writing the equations of velocity is for the two bodies after a specific time and equating their product to 0 with can solve the problem as shown below.


Problem

An object is thrown horizontally from a point it hits the ground at some another point. The line of sight from these two points makes an angle 60° with the horizontal. If acceleration due to gravity is 10 m/s Squire and time of flight is known what is the velocity of projection?

Solution

by writing the equations were displacement along x-axis and y-axis and by using a trigonometrical definition we can solve the problem as shown below.


Problem

Two paper screens are separated by a distance of hundred meter. A bullet fired through them. The hole made in the second screen is 10 cm below the hole made in the first screen. What is its initial velocity before it strikes the first screen?

Solution

Initially the bullet is having only horizontal velocity and its vertical component of velocity equal to 0. There is no impact of acceleration due to gravity along the x-axis and it is acting along the y-axis. Taking this things into consideration we can equate the distance between the two poles along the vertical direction is equal to the vertical distance travelled by the body during the journey time. We can also calculate the time taken by the body for this to happen using the equation of the displacement along the x-axis. By solving these equations together we can calculate the initial velocity of the projection as shown below.


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