IIT JEE and NEET Physics @ Venkats Academy

Showing posts with label Frequency. 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.

Time Period of Simple pendulum

Problems on Simple Pendulum with Solutions

Loaded spring in simple harmonic motion

Damped Oscillations and Forced Oscillation

Oscillations Problems with Solution One

Oscillations Problems with Solution Two

Units and Dimensions an Introduction Video Lesson

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.

Simple Harmonic Motion Displacement ,Velocity and Acceleration

Time Period,Frequency and Phase of a body in SHM

Problems on SHM displacement,velocity and Acceleration

Energies of Body in Simple Harmonic Motion

Oscillations Problems with Solution One

Oscillations Problems with Solution Three

Time Period of Simple pendulum

Problems on Simple Pendulum with Solutions

Loaded spring in simple harmonic motion

Damped Oscillations and Forced Oscillations

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.

Simple Harmonic Motion Displacement ,Velocity and Acceleration

Time Period,Frequency and Phase of a body in SHM

Problems on SHM displacement,velocity and Acceleration

Energies of Body in Simple Harmonic Motion

Thermodynamics Problems with Solutions Five

Thermodynamics Problems with Solutions One

Thermodynamics Problems with Solutions Two

Thermodynamics Problems with Solutions Three

Doppler Effect and Its Applications

The apparent change in the frequency due to the relative motion of source and observer is called Doppler Effect. We can experience the change in the frequency only when there is a relative motion. The original frequency of the source is not actually changing. Due to the relative motion it is appearing like changing and that’s why it is called as apparent change in frequency.

We can derive the equation for the apparent frequency in different possible cases. When the observer is in the motion he will receive more number of the waves than when he is in the state of rest. It is simply because waves are not only crossing him and he is also crossing the waves.

When the observer is crossing the stationary source there will be difference in the frequencies. If the observer is approaching the apparent frequency increases and when the observer is receding the apparent frequency decreases. The difference between these frequencies can be heard like beats to the observer. We can calculate the number of the beats as shown below.

There will be apparent change in the frequency even when the observer is the state of rest and the source is moving towards the Observer. Here is the source is approaching the observer, its wavelength towards the observer decreases and hence frequency increases. We can derive the equation for the apparent frequency in this case as shown below.

When the observer is in the state of the rest and source is approaching him, apparent frequency increases. When the source is moving away from the stationary observer, apparent frequency decreases. The difference between these two frequencies can be heard like the beats to the observer. We can derive the equation for the number of the beats as shown below.

When a source is revolving around the stationary observer, he is not going to have any Doppler effect experience. It is simply because there is no relative motion between the source and observer. No component of the velocity of the source is acting towards the observer and hence we cannot find any change in the frequency.

If the source is moving in the circular path and observer is far away from the Centre of the circular path, he can hear apparent frequency with different possible frequencies. When the source is moving away from the observer, apparent frequency decreases and vice versa. We can write the equation is as shown below.

When a source is moving by making an angle to the direction of the observer, still there will be apparent change in the frequency due to the component of the velocity of the source towards Observer. We can write the equation for it as shown below.

Problem and solution

Heat and Thermodynamics

Thermal Convection and Radiation

Newton’s law of cooling

Wave Motion an introduction

Progressive Wave Representation and Problems

Frequency of Stationary wave in a strecthed String

Stretched String Problems and Solutions

Vibrations of Sound in Closed Pipe and Open Pipe

Beats and Its Applications

The phenomenon in which to sound notes of slightly different frequency travelling together in the same direction are superimposed and produce alternate waxing and waning s called beats.

When the two waves are met in the same phase, they produces a maximum resultant intensity and it is called waxing. When the two waves are met in the opposite phase, they producers minimum resultant intensity and it is called waning.

We have derived a mathematical equation for beat frequency here.It is number of beats per one second.The time taken for completion of one beat that is one waxing and one waning is called time period of beat.A mathematical equation is derived for both of them as shown below.

The time interval between two maximum intensities as well as the two minimum intensities is always fixed. This is called Beat time period. The reciprocal of this time period is called beat frequency. We can derive the equation for them as shown below.

The diagrammatic view of the phenomena is as shown below.We can see one two waves are met in same phase,their resultant intensity is maximum and it is called waxing.As the time progresses,the phase difference increases and minimum intensity is produced and it is called waning.

The interval between two waxings and wanings is regular and systamatic.

Every ordinary human being needs a time interval of 0.1 second between the two successes sounds to understand the sound properly. This is called persistence of hearing. Hence difference between the frequencies of two sources shall not be greater than 10 to hear the beats.

Problem and solution

A tuning fork A has a frequency 5% more than the standard fork K and another tuning fork B has a frequency 3% less than the standard fork K. When this two tuning forks are vibrated together calculate the number of the beats generated?

Number of the beats generated is equal to difference between the frequencies.We can solve the problem as shown below.

Problem and solution

Solving problems in the concepts of beat is very simple.Just follow the concept given and comment if any clarification is required.