Oscillations

The general equation for simple harmonic equation is given as,

Now comparing this to the given equation we can say that,

Where T is the time period.

Thus, we see option (b) is correct.

We know that,

Thus, we can write,

Now differentiating the above twice we get,

But is not proportional to

Thus, it is not a SHM, but since there is a sine term, thus the motion will be periodic.

And , thus

Hence option (b) is correct.

For SHM,

∴

Thus, option (d) is correct

Imagine a u-tube having water to a certain level equal on both the sides. Now if the force F is applied on the one side of the u-tube compressing the liquid by and raising the liquid in by in the oher arm.

Thus, the restoring force due to the change in height is given by,

Where is density of the liquid and is the area of cross section.

Thus force is proportional to

Thus, it is a simple harmonic motion.

But

Where height is h

We see that time period is independent of density of liquid.

Thus, option (c) is correct.

and

Squaring and adding the above two we get,

But

Thus,

∴ the equation of motion is a circle of radius , thus option (c) is correct.

Consider the equation,

now, let and

now putting the above in the given equation,

Now using the identity

Thus, we get,

Thus, the motion is SHM with amplitude

∴ the option (d) is correct.

Time period is for SHM.

Therefore, on giving transverse displacement on a will make all vibrate but the length of a and c are same and thus they will be in resonance and thus c will have max amplitude.

Thus option (b) will be true.

At t=0 the position is , and T=30,

∴

At t=T/4 the position is and

Thus, the equation, satisfies the situation and thus option (a) is the correct option.

Now differentiating the above twice we get,

Thus, we find that is proportional to and hence it is an oscillatory.

Thus, the motion is not periodic.

Hence the option (c).

We know that acceleration

And velocity

∴

Thus,

On putting the values we get,

Hence option (a) is correct.

When the spring I connected individually,

Frequency of S₁ and S₂,

------(1)

------(2)

Now when they are attached as shown in the figure, it is a parallel connection and thus

Putting equation 1 and 2 in the above, we get,

Hence option (b) is correct.

Earth completes one rotation around its axis in 24 hrs and thus has a periodic motion but does not move to and fro about a mean position, thus, it does not have SHM.

Thus, option (a) and (c) are correct options.

When the ball bearing is released from a point slightly above the lower point the only restoring force is the force due to gravity.

Thus, when the displacement is ,

And the displacement is very small, thus,

Thus, a is proportional to -θ

∴ the motion is simple harmonic and periodic, so option a and c are correct.

In option (a), the state at t=0 and t=2 are not the same as the displacement are not equal. Thus incorrect.

In option (b), the state at t=2s and t=6s are in the same state as the displacement and direction are the same. Thus correct.

In option (c), though the displacement is same at t=1s and t=7s but the direction is opposite. Thus incorrect.

In option (d), the state at t=1s and t=5s are in the same state as the displacement and direction are the same. Thus correct.

∴ the option (b) and (d) are the correct.

For SHM to take place force must be proportional to displacement and opposite to it.

Consider the SHM equation for displacement,

Differentiating above w.r.t time t,

∴ velocity is periodic.

Differentiating once again w.r.t time, t,

Thus,

acceleration is not constant but periodic.

Thus, option (a), (b) and (d) are correct.

From the graph we can write the equation of SHM as,

And

Force

For

Thus option (a) is correct.

For maximum acceleration,

Thus

∴ option (b) is correct,

For max velocity

Thus option (c) is correct.

Kinetic energy and Potential

Now, if

Thus,

Thus, option (d) is wrong.

Consider SHM,

Where A is the amplitude and is the angular frequency

Now velocity is given as,

Max velocity is given as, ----(1)

mean velocity ---(2)

square mean velocity

solving this we get, ---(3)

thus from equation (2) we can say that option c is incorrect and from equation (3) we can say that option (d) is correct

Now the instantaneous potential energy,

Where

---(4)

Total kinetic energy in SHM s given as,

---(5)

Average kinetic energy

Solving the above we get, ---(6)

Similarly,

Average potential energy

Solving the above we get, ----(7)

And average total energy

---(8)

Max kinetic energy ----(9)

And max potential energy ---(10)

Comparing equation (8) and (9) we see option a is correct.

Comparing equation (9) and (6) we see option b is also correct.

The force always acts in the direction of the mean position and at point C it is acting towards O and away from A. Thus positive.

Also, acceleration is in the direction of force thus it is also acting in the same direction as force and is positive.

Also, the direction of motion is towards A and thus velocity is also positive.

Hence option (a) is correct.

b) when particle is at C the velocity is in direction towards O thus positive and hence option (b) is incorrect.

c) when the particle is at 4 cm away from B and moving towards The acceleration and force are directed towards mean position O and also since it is moving towards A then all velocity, acceleration and force are negative in signs. Option c) is correct.

d) acceleration and force is always towards mean position O and thus at B the acceleration and force is negative. Thus correct.

The maximum velocity is always at the mean position of the SHM where displacement is zero

Thus, we can say that points B, D, F, H are the points where the velocity is maximum.

The minimum velocity is always at the points where the displacement is always maximum.

Thus points A, C, E, G are the points of where velocity is zero.

Consider the direction towards k to be negative and towards K to be positive.

Now let the box be displaced by x towards spring k and thus the displacement =

Now the force due to spring k is and is directed towards right side due to compression of the spring

and the force due to spring k is and is directed towards right side due to extension of the spring

∴ total restoring force

Two basic characteristics of a simple harmonic motion are:

i) Restoring force is proportional to negative of the displacement.

ii) Acceleration is always directed towards the mean position.

Forces acting on the pendulum,

Tension T due to the string

Force of gravity due to mass of pendulum in vertically downward direction, mg

Now the force of tension T is balanced by the cosine component of the weight, i.e.,

And the sine component of the weight is called the restoring force acting towards the equilibrium position of the pendulum

, now if θ is very small then, sin θ θ and we can write for θ ,

, where L is the length of the pendulum and is the tangential displacement.

∴

Clearly F is proportional to the displacement and in opposite direction of tangential displacement

When the pendulum is displaced by a small angle θ such that the sin θ≅ θ, then the pendulum will execute simple harmonic motion.

Equation of motion for S.H.M is as follows,

Where,

x is displacement

A is the amplitude

w is the angular velocity

t is the time

for velocity, we differentiate the above equation, we get,

For max v,

----(1)

for acceleration, we differentiate the velocity equation, we get,

For max a,

----(2)

Dividing equation 2 with 1 we get,

∴ the ratio of max acceleration to max velocity is

Motion of oscillator can be divided into 4 parts,

At t=0, it is at its mean position

At t=T/4, it covers the maximum distance A

At t=T/2, it moves back to its mean position from maximum amplitude and covers a distance of A

At t=3T/4, it again moves to its maximum position and covers a distance of A

At t=T, it again moves back to its mean position and covers distance of A

∴ in one cycle it covers total of 4A distance

∴ ratio of the distance covered in once cycle to the amplitude

(Note: in this question we are asked to find the distance and not displacement ∴ we do not use S.H.M. equation , which gives displacement of the particle from its mean position)

As the particle P moves in anti-clockwise direction, its x-axis projection length decreases ∴ the distance OP’ decreases and this is only possible if P’ moves towards negative x-axis direction.

Hence we can say that the sign of the velocity of P’ will be negative

For S.H.M we can write the S.H.M equation as follows,

----(1)

Where x=displacement

w=angular velocity

t = time

for velocity we differentiate the above equation,

From trigonometric property we know that

∴ ----(2)

Hence comparing equation (1) and (2) we can say velocity and displacement have a phase difference of .

Total energy of S.H.M is constant, ∴ we get a straight line

Kinetic energy for the simple harmonic motion is given as,

where x = displacement

ω = angular velocity

m = mass of the system

a = amplitude

the above equation is an equation of the downward opening parabola symmetric along y-axis and vertex at on y-axis

Potential energy for the simple harmonic motion is given as,

where x = displacement

ω = angular velocity

m= mass of the system

a = amplitude

the above equation is an equation of the upward opening parabola symmetric along y-axis and vertex at (0,0) on y-axis

∴ following is the graph we get,

Time period of a second’s pendulum is,

, where T = time period

L = Length of the pendulum

g = acceleration due to gravity

∴

For moon,

------(1)

For earth,

------(2)

Dividing equation 1 by 2 we get,

∴

Therefore, the length of the second pendulum will be 170 cm

Let the spring be extended by when loaded with mass m.

Due to extension there will be a restoring force applied by the spring on the pulley

Now let T be the tension on the pulley,

∴ where k’ is the spring constant.

pulley experiences two tension force due the two strings,

∴we get,

-----(1)

Where k is the spring constant.

Now let us pull the mass by , as we pull it by the spring extends by a distance of 2 because the string on one side is inextensible.

∴ the new tension force

pulley experiences two tension force due the two strings,

∴the net force acting,

-----(2)

Now putting equation (1) in (2) we get,

∴ we see that net force F is proportional to the the system performs a simple harmonic motion.

Comparing the above obtained net force with

We find that,

And for S.H.M the time period T is given as,

Hence the time period for the given system is,

Given:

Multiplying both the sides by

Writing we get,

Using the identities in equation above,

We get,

∴ angular velocity from above equation

∴ time period

Potential energy for the simple harmonic motion is given as,

And the total energy,

where x = displacement

w= angular velocity

m= mass of the system

A = amplitude

Now we are given that,

⇒

⇒

∴ the displacement will be units

We know that the derivative of the potential energy gives the force, i.e.,

-----(1)

We are given that,

∴

Putting the above in equation (1) we get

For very small we get,

----(2)

And for simple harmonic motion we know that, -----(3)

Where k is spring constant

Comparing equation 2 and 3

For simple harmonic motion time period is given as follows,

, where k is the spring constant and m is the mass of the system and T is the time period

∴ putting k in the time period equation, we get,

For simple harmonic motion, the displacement equation is given as follows,

----(1)

where x = displacement

w= angular velocity

A = amplitude

We will be finding each parameter from the given data and then substitute in the above equation

Given:

mass m

spring constant

amplitude ------(2)

time period for S.H.M.

Putting the values, we get,

Now we need angular velocity

Which is written as,

∴ ----(3)

Putting (2) and (3) in (1) we get,

the expression for its displacement at any time t is given as

For simple harmonic motion,

where x = displacement

w= angular velocity

A = amplitude

δ = phase

Given:

for both the pendulum the amplitude is given to same and ∵ they are identical, their length are the same and thus their angular velocity are also same as for the pendulum , g=acceleration due to gravity and length of the pendulum is given as L.

∴ pendulum 1 we can write,

At

∴

-----(1)

pendulum 1 we can write,

At

∴

-----(2)

For phase difference we subtract (2) from (1), we get,

∴ the phase difference between the two pendulum is of

Given:

Mass = 50kg

Harmonic frequency = 2.0

∴ angular velocity

Amplitude A = 5.0 cm

The weighing machine will measure the normal reaction of the body as weight.

Now let be the acceleration of the platform

When the platform raises up,

From the above free body diagram we get the following force balance equation,

-----(1)

where N is the normal reaction

which shows that the weight measured by the machine will be less than the actual, since the machine shows the normal reaction and not mg

When the platform moves down,

From the above free body diagram we get the following force balance equation,

-----(2)

where N is the normal reaction

which shows that the weight measured by the machine will be more than the actual, since the machine shows the normal reaction and not mg

a) In the cases when platform rises and drops the weight measured are different. ∴ answer to part (a) is yes

b) Since the reading by the machine depends on the Normal reaction. Therefore, it will be maximum and minimum when the normal reaction is maximum and minimum respectively.

For minimum reading we put given values in equation (1),

For maximum reading we put given values in equation (1),

a) When the box falls its potential energy changes and the change in potential energy = energy stored in the spring

∴

Where k=spring constant, x=displacement and m=mass of the box

-----(1)

Now at the equilibrium, let the amplitude of the spring be

∴ balancing the forces,

------(2)

Putting (1) in (2), we get,

Now

∴ the amplitude of oscillation = 2 cm

b) Time period of oscillation of the system is given as,

Now from equation 1, we have,

Putting the above in the frequency equation we get,

Hz

∴ the frequency is 3.52 Hz

Here we apply the concept of buoyant force

Let the wood part of the height of the wood be immersed into the water

Here the wood log experiences a buoyant force which is given as,

,

where ρ =density of wood log, V=volume of the water displaced by wood now at equilibrium, the buoyant force is balanced by the weight, i.e.,

------(1)

Now let the wood be displaced by depth into the water,

∴ the volume of the water displaced by the wood is,

∴ the buoyant force

∴ there acts net restoring force on the wood log,

Putting equation (1) in the above we get,

-------(2)

∵ is proportional to

∴ wood performs a simple harmonic motion

Now the in S.H.M., we have

Comparing this equation with equation (2) we get,

For S.H.M, Time period

Substituting the value of k, we get,

Hence proved.

In this problem we prove the S.H.M. by deriving standard equation ------(1),

which represents simple harmonic equation.

Let the height of the fluid in the left and right column are respectively.

Consider an element with mass which is at a height of unit from the bottom in the left column can be related as follows,

, where ρ = density of the fluid in the tube

A=Area of cross-section of the tube

For total potential energy we integrate the above,

Similarly, for right column we have,

∴ total potential energy,

Now initially

Let the vertical height of the liquid on both the tube be and when the pressure difference is produced the liquid in one tube drops by and liquid in the other raises by height of .

New height at left column for the fluid =

And new height at left column for the fluid =

∴ the total

Now the change in potential energy

Change in kinetic energy, ----(2)

Now if is the length of the tube,

then

And mass m is given as

Putting the above in the equation (2), we get,

∴ total energy =

Now the change in total energy is =0

∴

Differentiating w.r.t. time t,

()

(∵ acceleration and here )

Now putting , we get,

-------(3)

Comparing (1) and (3) we see that the above equation represents S.H.M. and on comparing we get,

When the body is made to fall through the tunnel it falls with some acceleration, and on reaching the centre it has some velocity to move ahead. ∴ the velocity vector points away from the centre. But gravity of the earth tries to pull it back due to which it experiences force in opposite direction of motion. All happens when it crosses the centre for the first time.

And acceleration due to gravity inside the earth is given as,

,

where d=depth from the surface of the earth

and R=radius of the earth

∴ net force

Now let , which the distance from the centre of the earth.

∴ we see that is proportional to -

Hence it is Simple Harmonic Motion.

Comparing the above equation with ,where k is the spring constant,

We get,

And time period

The angular displacement for the pendulum is given as,

----(1)

Where is angular displacement at the given time t and is the maximum angular displacement of the pendulum, and is the angular frequency and t is the time t.

Given:

Time period T of oscillation is = 1s

Length of the pendulum is =

∴

At ,

and,

at the string snaps,

putting the above in equation (1)

now since pendulum moves in the circular motion it must have angular velocity. For that, we differentiate eq. (1)

But

∴

At t=1/6 s

------(2)

But we have a following relation between angular and linear velocity,

, where v is the linear velocity and r is radius and in this case r is the length of the string of the pendulum

∴

Putting the above in equation (2)

We get,

------(3)

∴ the linear velocity at t=1/6 s is given as above.

Now at this stage the string snaps and bob starts a parabolic motion with the above calculated velocity in the tangential direction, thus we need to resolve it into x and y axis component.

Let us look at the vertical component,

Let the bob be at a height from the ground

Velocity along vertical axis,

Now using 2^nd equation of motion, as the bob falls by H₀ distance,

Now solving for t in the above quadratic equation we get,

Now neglecting the term containing , we get,

This is the same time period for which the bob travels along the x-axis after snapping takes place.

And there is no acceleration in the x-axis we can use following equation,

------(4)

Where s is the distance travelled along the x-axis and using equation (3) we can write,

(given )

∴ putting it in equation (4)

And

Since we are given that we can take =1

We get,

∴

Now the horizontal distance from the point A at the time it snaps is,

(since is very)

∴ the distance at which it falls from the point A should be as follows,

The above is the final answer.