Oscillations

A to and fro motion about a mean position having a definite period is a periodic motion.

Let us analyse each condition-

(a) Here, the motion of the swimmer between the banks of the river is to and fro. But it cannot be said to have a definite period as we do not know how much time the swimmer will need to complete each trip. Hence, it does not represent a periodic motion.

(b) When a freely suspended magnet is displaced from N-S direction and released, it oscillates periodically about the mean position. Hence, it is a periodic motion.

(c) The hydrogen molecule rotates about its centre of mass which constitutes a periodic motion.

(d) When an arrow is released from a bow, it just goes in one direction and doesn’t make oscillations back and forth. Hence, the motion is non-periodic.

(a) The rotation of the earth about its axis is a periodic motion(it has a period of 24 hours). But as it is not to and fro motion, it does not represent S.H.M.

(b) The mercury column which oscillates back and forth in a U-tube represents S.H.M.

(c) The ball bearing represents S.H.M as it oscillates to and fro in the bowl.

(d) The general vibrations of polyatomic molecule about its equilibrium position are periodic. But they do not represent S.H.M as it is a superposition of SHMs by individual atoms.

A periodic motion is the one performed by particles oscillating back and forth at regular intervals.

Here, in x-t plots it is seen,

(a) As the time proceeds, the particle continues to proceed with increasing distance without oscillating. Hence, the motion is non-periodic. (Refer Fig 14.23-(a))

(b) The particle vibrates back and forth continuously, producing a periodic motion with a time period of 2 seconds. (Refer Fig 14.23-(b))

(c) The motion is non-periodic as seen in the graph. (Refer Fig 14.23-(c))

(d) The motion is periodic with a period of 2 seconds. (Refer Fig 14.23-(d))

The condition for a function to be periodic is that it must identically repeat itself after a fixed interval of time. For a function to represent an SHM, it must have the form of cos ( or sin ( with a time period T.

(a) sin ωt – cos ωt = √2(-)..(Multiply & divide by √2 )

= (sin ωt.cosπ/4 - cos ωt.sinπ/4)

= √2.sin (ωt - π/4)

Hence, it is an SHM with a period 2π/ω.

(b) sin³ ωt = 1/3(3sin ωt - sin3ωt)

Each term here, sin ωt and 3ωt represent SHM. But B. is the result of superposition of two SHMs, is only periodic not SHM. Its time period is 2π/ω.

(c) It can be seen that it represents an SHM with a time period of 2π/2ω.

(d) It represents periodic motion but not SHM. Its time period is 2π/ω.

(e) An exponential function never repeats itself. Hence, it is a non periodic motion.

(f) It clearly represents a non-periodic motion.

Let us analyse each case: Refer the figure above.

(a) At the extreme position A, the velocity is zero, acceleration and force are directed towards the mid point O and are positive.

(b) At the end B, another extreme position, velocity is zero, acceleration and force are directed towards O and are negative.

(c) Here, the velocity is negative and maximum. Acceleration and force are zero.

(d) Velocity, acceleration and force are negative as they travel in the negative direction.

(e) Velocity is positive. Acceleration and force are directed towards O and are positive.

(f) Velocity is directed along BA, hence it is positive. Acceleration and force are directed towards OB and are positive.

The relation between displacement and acceleration involves S.H.M when acceleration is proportional to displacement and is directed in the opposite direction.

In (c) acceleration is proportional to displacement x and directed in the opposite direction. Hence, only (c) a = –10x involves Simple Harmonic motion.

We have the given displacement function as

x(t) = A cos (ωt + φ ) …….(1)

The initial conditions are,

t = 0, x(0) = 1 cm i.e the position of the particle and angular frequency, ω = π s^–1

Substituting the initial conditions in the displacement function in equation(1)

Hence, 1 = A cos(π .0+ φ )

A cos φ = 1 ………(2)

Differentiating (1) w.r.t ‘t’

The derivative of displacement is velocity.

Hence,

v = -Aωsin(ωt + φ ) …………(3)

At t = 0, v = ω (Initially)

Hence, from (3) we get,

ω = -Aωsin(π .0 + φ )

Also, Asinφ = -1 …………….(4)

Now we square and add (1) and (4)

We get A = √2 cm

Dividing (2) and (4),

Asin φ/Acos φ = -1/1

Hence, φ = 3π/4

We use the sine function

x = B sin (ωt + α)

we get Bsin α = 1 …..(5)

at t = 0, using x = 1 and v = ω

and Bsin α = 1 ……….(6)

Dividing (5) and (6)

tan α = 1 and α = π/4 and 5π/4

Squaring and adding (5) and (6) we get

B = √2 cm.

Given

Mass of the spring balance,M = 50 kg,

Length of the scale,Y = 20 cm = 0.2 m,

Period of oscillation,T = 0.60 seconds.

We know,

F = ky or M = ky i.e

Mass ×acceleration due to gravity = Spring constant × Length of scale.

Hence, k = Mg/0.2

= 50×9.8/0.2 N/m

= 2450 N/m

Now, T = 2π√m/k

T² = 4π²m/k

Hence, m = T²k/4π²

Substituting the values we get,

m = kg = 22.3 kg

Hence, mg = 218.5 N = 22.3 kgf.

Here, Spring Constant,K = 1200N/m

Mass,m = 3 kg

Distance travelled by mass,a = 2 cm = 0.2 m

(i) Frequency, v = 1/T = 1/2π√k/m

= 1/2×3.14√1200/3

= 3.2 /s

(ii) Acceleration, A = (k/m)y

When y = a i.e maximum, acceleration will be maximum

Hence,A_max = (ka)/m

= (1200×0.02)/3

= 8 m/s²

(iii) When the mass will be passing through mean position, it will have maximum speed.

V_max = a√(k/m) = 0.02×√(1200/3)

= 0.4 m/s.

Distance,a = 2 cm,

ω = √k/m

= √1200/3 s^-1

= 20 s^-1

a) As the time is measured from mean position, the phase is 0.

x = a sinωt = 2 sin 20t

b) At the maximum stretched position, the body will be at the extreme right position. The initial phase is π/2.

x = asin (ωt+ π/2) = a cos ωt = 2 cos 20t

c) At the maximum compressed position, the body will be at the extreme left position. The initial phase is 3π/2.

x = asin (ωt+ 3π/2) = -a cos ωt = -2 cos 20t

We know as a particle move in circular path the projection on x and y axis of displacement covered by particle represent a simple harmonic motion as the particle moves the angle subtended by it changes and if velocity is uniform its angular velocity or rate of change of angle subtended at centre with respect to original position become constant (a fixed value), projections can be represented as a function of sine and cosine in terms of angular velocity or time period and time

(a) Here particle started from a point on negative Y axis and started moving in a circular path in clockwise direction, now the time period of one complete rotation is given as T = 2s, so the particle will return to its original position in 2s, so after 1 second it will be on diametrically opposite point

Now let angle made by line joining particle to centre make an angle 𝜽 with respect to negative y axis, 𝜽 will vary with time, now at any instant a triangle is formed

As shown in figure

Here suppose particle is at A and angle made between negative y axis and line joining its position to centre of circle is 𝜽 at any instant of time t, now length AB is the projection on x axis at any instant now radius of circular path 3 cm so maximum projection on x axis can be 3 cm or amplitude can be 3 cm

Now let the angular velocity of motion be ω, now angular velocity can be calculated with the help of total time period on one complete revolution

We know relation between time period T and angular frequency ω ω = 2π/T

Here time period is

T = 2s

So angular velocity is

ω = 2π/2 = π rad/s

now angle between negative y axis and line joining its position to centre of circle is 𝜽 at any instant of time t will be

𝜽 = ωt

Now from the triangle AOB we can see it is a right angled triangle so using trigonometry we have

Sin𝜽 = AB/OA

Or we can say

AB = OA Sin𝜽

now let the projection on x axis as

AB = -x

(since distance along negative x axis)

𝜽 = ωt

i.e. 𝜽 = πt

we know OA equal radius of circle so we have

OA = 3 cm

So we have

-x = 3 Sin πt

Or

The projection of displacement of particle on x axis at any instant of time can be represented by equation

x = -3 Sin πt cm

which is the equation of a simple harmonic motion with angular velocity π rad/s and amplitude 3 cm

(b) Here particle started from a point on negative X axis and started moving in a circular path in anticlockwise direction, now the time period of one complete rotation is given as T = 4s, so the particle will return to its original position in 4s, so after 2 second it will be on diametrically opposite point

Now let angle made by line joining particle to centre make an angle 𝜽 with respect to negative x axis, 𝜽 will vary with time, now at any instant a triangle is formed

As shown in figure

Here suppose particle is at B and angle made between negative X axis and line joining its position to centre of circle is 𝜽 at any instant of time t, now length 0A is the projection on x axis at any instant now radius of circular path 2m so maximum projection on x axis can 2 m or amplitude will be 2 m

Now let the angular velocity of motion be ω, now angular velocity can be calculated with the help of total time period on one complete revolution

We know relation between time period T and angular frequency ω = 2π/T

Here time period is

T = 4s

So angular velocity is

ω = 2π/4 = π/2 rad/s

now angle between negative x axis and line joining its position to centre of circle is 𝜽 at any instant of time t will be

𝜽 = ωt

Now from the triangle AOB we can see it is a right angled triangle so using trigonometry we have

Cos 𝜽 = OA/OB

Or we can say

OA = OB Cos𝜽

now let the projection on x axis as

OA = -x

(since distance along negative x axis)

𝜽 = ωt

i.e. 𝜽 = π/2 t

we know OB equal radius of circle so we have

OB = 2 m

So we have

-x = 2 Cos π/2 t

Or

The projection of displacement of particle on x axis at any instant of time can be represented by equation

x = (-2 Cos π/2 t)m

which is the equation of a simple harmonic motion with angular velocity π/2 rad/s and amplitude 2 m

We know here we are given equation of simple harmonic motion as

x = –2 sin (3t + π/3)

Converting to Cosine

Using Cos(π/2+𝜽) = -Sin𝜽

We get

x = –2 sin (3t + π/3) = 2 Cos (3t + π/3 + π/2)

or we get

x = 2 cos (3t + 5π/6)

now comparing it with standard equation of S.H.M

x = A cos(ωt + ϕ)

where x is the position of particle in time t moving in S.H.M. with angular frequency ω and ϕ is the initial phase angle,A is the amplitude

clearly since here coefficient of cosine term is 2 so amplitude is 2 cm i.e radius of circle will be 2 cm

now initial position of particle or position at t = 0 sec will be

x = -2 sin (0 + π/3) = -2Sin(π/3)

i.e. (since )

i.e x component of displacement of particle is in negative x direction

the coefficient of t is angular velocity so here the angular velocity is

ω = 3 rad/s

we have to assume particle to be moving in anticlockwise direction

and initial phase angle is

ϕ = 5π/6 = 150⁰

i.e. line joining centre of circle to position of particle makes angle of 150⁰ with x axis in anticlockwise sense

so the plot is as shown in figure

we know here we are given equation of simple harmonic motion as

x = cos (π/6 – t)

rearranging equation

x = 1cos (1t - π/6)

(using Cos (-𝜽) = Cos𝜽)

now comparing it with standard equation of S.H.M

x = A cos(ωt + ϕ)

where x is the position of particle in time t moving in S.H.M. with angular frequency ω and ϕ is the initial phase angle,A is the amplitude

clearly since here coefficient of cosine term is 1 so amplitude is 1 cm i.e radius of circle will be 1cm

now initial position of particle or position at t = 0 sec will be

x = cos (0-π/6) = cos (-π/6)

i.e. x component of displacement or projection on x axis is

the coefficient of t is angular velocity so here the angular velocity is

ω = 1 rad/s

we have to assume particle to be moving in anticlockwise direction

and initial phase angle is

ϕ = -π/6 = 30⁰

i.e. line joining centre of circle to position of particle makes angle of 30⁰ with x axis in clockwise sense

so the plot is as shown in figure

We know here we are given equation of simple harmonic motion as

x = 3 sin (2πt + π/4)

Converting to Cosine

Using Cos(𝜽 + π/2) = Sin𝜽

We get

x = 3 cos (2πt + π/4 + π/2) = 3 Cos (2πt + 3π/4)

or we get

x = 3 Cos (2πt - π/4)

now comparing it with standard equation of S.H.M

x = A cos(ωt + ϕ)

where x is the position of particle in time t moving in S.H.M. with angular frequency ω and ϕ is the initial phase angle, A is the amplitude

clearly since here coefficient of cosine term is 3 so amplitude is 3 cm i.e radius of circle will be 3 cm

now initial position of particle or position at t = 0 sec will be

x = 3 sin (0 + π/4) = 3Sin(π/4)

i.e. (since )

i.e x component of displacement of particle is in positive x direction

the coefficient of t is angular velocity so here the angular velocity is

ω = 2π rad/s

we have to assume particle to be moving in anticlockwise direction

and initial phase angle is

ϕ = -π/4 = -45⁰

i.e. line joining centre of circle to position of particle makes angle of 45⁰ with x axis in clockwise sense

so the plot is as shown in figure

We know here we are given equation of simple harmonic motion as

x = 2cos (πt)

now comparing it with standard equation of S.H.M

x = A cos(ωt + ϕ)

where x is the position of particle in time t moving in S.H.M. with angular frequency ω and ϕ is the initial phase angle,A is the amplitude

clearly since here coefficient of cosine term is 2 so amplitude is 2 cm i.e radius of circle will be 2cm

now initial position of particle or position at t = 0 sec will be

x = 2cos (0) = 2

i.e. x component of displacement or projection on x axis is

x = 2 cm

the coefficient of t is angular velocity so here the angular velocity is

ω = π rad/s

we have to assume particle to be moving in anticlockwise direction

and initial phase angle is

ϕ = 0 rad = 0⁰

i.e. line joining centre of circle to position of particle makes angle of 0⁰ with x axis or is parallel to x axis

so the plot is as shown in figure

(a) Now here when springs are stretched by a force F they will apply a restoring force in opposite direction when extension in length of spring is there as the extension in the spring increases the restoring force in the spring also increases so there are two Forces acting on each block in opposite direction when external applied force is equal to the restoring force of the block the block will be in equilibrium the free body diagram for block in fig (a)

has been shown in the figure below

Restoring Force of spring F_s is acting in left direction and is equal in magnitude to applied force acting in right direction i.e. F = F_s

We know restoring force in a spring is given as

F_s = Kx

Where K is the spring constant of the spring and x ia the extension in the spring

So we have

F = Kx

i.e. x = F/K

or the extension in length of spring is

x = F/K

now in case (b) the spring is holding two blocks in equilibrium as spring always applies restoring force in a direction opposite to the direction of applied force, so for Block A on left external force is to the left so spring force will be in right direction and will be equal in magnitude to the external force and for Block B on Right external force is to the Right so spring force will be in Left direction and will be equal in magnitude to the external force, the free body diagram of both the blocks has been

shown in the figure

So we have

F = F_s (For Both the blocks)

We know restoring force in a spring is given as

F_s = Kx

Where K is the spring constant of the spring and x is the extension in the spring

So we have

F = Kx

i.e. x = F/K

or the extension in the length of spring is

x = F/K

(b) When the mass in (a) is releases it will start oscillating and perform simple harmonic motion since for we know condition for simple harmonic motion is that acceleration of particle is proportional to distance from mean position and is directed toward mean position

and we have relation between acceleration and displacement as

a = -ω²x

where a is the acceleration of particle undergoing Simple Harmonic Motion with angular frequency ω and x is the displacement from mean position

here on the block of mass m, when it is at a distance x from original position or mean position

force is

F = -Kx (Force is in opposite direction to displacement )

Where K is Spring constant

Now we know

F = ma

Where F is the force acting on a body of mass m whose acceleration is a

So we have

ma = -Kx

or the acceleration of the mass is

a = -(K/m)x

since spring constant K and mass of block m are constant quantities so we can say acceleration of block is proportional to displacement from mean position and we know it is directed towards mean position so block is executing simple harmonic motion so comparing with equation of simple harmonic motion we get

ω²= K/m

where ω is the angular frequency of simple harmonic motion

now we get

We know relation between time period T and angular frequency ω as

T = 2π/ω

So we have the time period of the oscillation as

So the time period of simple harmonic oscillation is

Where m is the mass of the block and K is the spring constant

Now in Case (b) again both blocks will move to and fro from each other and start oscillating simple harmonically and the mean position will be different for both blocks on either side of centre of spring

As shown in the figure

Now the spring can be divided in two parts and we can x₀ is the natural length of both part of spring on either side, now considering any one Block we can say time period of oscillation is

Where m is the mass of the block and K’ is the spring constant for that part

When a spring is cut to half value of spring constant doubles

So if K is the original spring constant, Spring constant of each part is

K’ = 2K

So time period is

Where m is mass of each block and K is the spring constant of the spring.

Now here we are given a piston in the cylinder head of a locomotive, the piston moves with simple harmonic motion and the stroke, or distance between two extremes of the piston is 1m

As shown in figure

Now the stroke is twice the amplitude of the simple harmonic motion so the amplitude is

A = 1/2 m = 0.5 m

Now were are given angular frequency of oscillation

ω = 200 rad/min

we know for a body undergoing simple harmonic motion the velocity is maximum when the body is at mean position or middle position and maximum velocity is given by relation

V = Aω

Where A is the amplitude and ω is the angular frequency of oscillation

So we get maximum velocity of piston as

V = 0.5m × 200rad/min

= 100 m/min

Now 1 minute = 60 second

So we have

V = 100m/60s = 1.67 m/s

So we get maximum velocity of the piston is 1.67 m/s

Time Period of a simple pendulum is time taken by pendulum to complete one complete oscillation, we know time period of a simple pendulum is given by the relation

Where T is the time period of the Simple Pendulum

l is the length of the pendulum, having a bob of mass m and g is the acceleration due to gravity

A simple Pendulum is shown in the figure

Let the time period of simple pendulum on earth be

Where T_e is the time period of the Simple Pendulum on earth

l is the length of the pendulum, and g_e is the acceleration due to gravity on earth

And let the time period of pendulum on moon be

Where T_m is the time period of the Simple Pendulum on Moon

l is the length of the pendulum, and g_m is the acceleration due to gravity on Moon

diving both equations we get

Solving and cancelling terms we get

Or we can say Time period of simple pendulum on moon is

Where Time period of simple pendulum on moon is

T_e = 3.5 s

Acceleration due to gravity on moon is

g_m = 1.7 m s^–2

Acceleration due to gravity on earth is

g_e = 9.8 m s^–2

so putting these value we get Time period of simple pendulum on moon is

So time period of the simple pendulum on moon is 8.40 s

In case of a simple pendulum the restoring force which is causing simple harmonic motion of bob of pendulum is

F = -mgSin𝜽

Where m is the mass of the Bob, g is acceleration due to gravity and 𝜽 is the angle made by string of pendulum with the vertical

As shown in the figure

We know force on a Body is

F = ma

Where m is the mass of particle and a is the acceleration of the particle, so acceleration of the body can be written as

a = F/m

so acceleration of bob is

a = -mgSin𝜽/m = -gSin𝜽

(negative sign because acceleration is in direction opposite to displacement)

when 𝜽 is very small we approximate value of sin𝜽 to 𝜽, for small oscillations

sin𝜽 ≈ 𝜽

acceleration of bob is

a = -g𝜽

further we know displacement can be expressed as

x = l𝜽

where x is the displacement of bob making an circular arc and 𝜽 is the angle covered and l is the length of pendulum so we get

𝜽 = x/l

And acceleration of pendulum as

a = -(g/l)x

in case of a mass attached to a spring

as shown in figure

suppose extension in spring is x so restoring force acting on mass will be

F = -Kx

We know force on a Body is

F = ma

Where m is the mass of particle and a is the acceleration of the particle, so acceleration of the body can be written as

a = F/m

so acceleration of mass is

a = -Kx/m

or a = -(K/m)x

now in both the cases acceleration is proportional to mass since in case of pendulum acceleration due to gravity g, length of pendulum is constant so acceleration of bob of pendulum is proportional to displacement from mean position and in case of spring, the spring constant K and mass m are constant so acceleration of mass is proportional to displacement of mass from mean position

so both are undergoing simple harmonic motion and we have relation between acceleration and displacement as

a = -ω²x

where a is the acceleration of Body undergoing Simple Harmonic Motion with angular frequency ω, x is displacement from equilibrium or mean position

so for pendulum we have

ω² = g/l

or angular frequency of oscillation as

We know relation between time period T and angular frequency ω T = 2π/ω

So we have the time period of the oscillation as

Now in case of mass attached to spring

ω² = K/m

or angular frequency of oscillation as

We know relation between time period T and angular frequency ω T = 2π/ω

So we have the time period of the oscillation as

As we can see in case of pendulum term of mass gets cancelled out but not so in case of a mass attached to a spring

In case of a simple pendulum the restoring force which is causing simple harmonic motion of bob of pendulum is

F = -mgSin𝜽

Where m is the mass of the Bob, g is acceleration due to gravity and 𝜽 is the angle made by string of pendulum with the vertical

As shown in the figure

We know force on a Body is

F = ma

Where m is the mass of particle and a is the acceleration of the particle, so acceleration of the body can be written as

a = F/m

so acceleration of bob is

a = -mgSin𝜽/m = -gSin𝜽

here we approximate for small oscillations

Sin𝜽≈𝜽

But for larger angles 𝜽 > sin𝜽

Further we express displacement as

x = l𝜽

where x is the displacement of bob making an circular arc and 𝜽 is the angle covered and l is the length of pendulum so we get

𝜽 = x/l

And acceleration of pendulum as

a = -(g/l)x

now without using approximation

we get angular frequency

So We know relation between time period T and angular frequency ω

T = 2π/ω

We get

so if we do not use the approximation the result for time period would not be same and actual time period is greater than calculated one, because the formula for time period

is valid when angle of oscillations are very small

Yes, the watch gives correct time as wrist watch works on spring oscillation whose Time period is given as

Where K is spring constant and m is mass

Clearly time period is independent of gravity in free fall effective gravity changes but here it will have no effect because time period of watch does not depend on gravity so it will keep on showing correct time

We know Frequency of a simple pendulum is given by the relation

Where 𝜈 is the frequency of oscillation, g is acceleration due to gravity, l is the length of the pendulum, in case of free fall effective acceleration due to gravity changes and becomes 0

g’ = 0

so putting g = 0 ms^-2 in the above equation

we get frequency of oscillation

So frequency of oscillation of pendulum in case of free fall is 0 Hz

Now when the pendulum will be in a car moving on a circular track of radius R with a uniform speed v, the effective acceleration on it will change which was earlier only acceleration due to gravity and now there is acceleration due to gravity and centripetal acceleration both acting perpendicular to each other so net acceleration and its direction also will change, the mean position will also change and the new mean position or equilibrium position of bob of pendulum will be in direction of acceleration as shown in the figure

Now centripetal acceleration on a body moving in circular path with uniform speed is given as

a_c = v²/R

where v is the uniform speed of the body and R is the radius of circular path

acceleration due to gravity g is acting in downward direction so net acceleration is

or

i.e.

we know time period of oscillation of a pendulum is given as

Where T is the time period l is the length of the pendulum and g is acceleration due to gravity

Since here now effective acceleration acting on pendulum has changed the equation will become

Where a is the effective acceleration acting on bob of pendulum

Putting value of a we get time period of oscillation of pendulum

Where T is the time period l is the length of the pendulum which is in a car which is moving on a circular track of radius R with a uniform speed v and g is acceleration due to gravity.

Now when a cork is kept on Liquid surface there are two forces acting on it equilibrium position (when forces are balanced and it is at rest), its weight acting in the downward direction and up thrust due to liquid which is equal to weight of liquid displaced by the block, now for equilibrium condition suppose A cylindrical piece of cork of density ρ of base area A and height h floats in a liquid of density ρ_l and its Y height is inside the liquid

As shown in the figure

Now volume of liquid displaced equal to the volume of cork inside liquid, now we know

Volume = Area × length

Here surface area of cork is A, and suppose length inside liquid is Y so the volume of cork inside liquid is

V₁ = AY

Now total height of cork is h so total volume of the cork is

V = Ah

Now we know mass is given as

m = V × ρ

where m is the mass of the body, V is its volume and ρ is the density

so mass of the cork is

m = Ahρ

where A is the area of cross section of cork, ρ is its density and h is its height

we know weight of a body is given as

W = mg

Here m is the mass of the body and g is acceleration due to gravity, so weight of the cork is

W = Ahρg

now mass of water displaced by the cork is

m₁ = AYρ_l

where A is the area of cross section of cork, ρ_l is its density and Y is its height inside liquid

we know weight of a body is given as

W = mg

Here m is the mass of the body and g is acceleration due to gravity, so weight of liquid displaced by the cork is

W₁ = AYρ_lg

Where ρ_l is the density of the liquid, Y is the height if cork inside liquid

Now force equal to weight of liquid displaced by cork will act in upward direction and weight of cork will be in down ward direction both must be equal in magnitude and opposite in direction for equilibrium i.e.

W = W₁

Now when the cork is pushed inside slightly the amount of liquid displaced by it will increase hence up thrust due to liquid will increase and but weight will remain same hence there will be a net force in upward direction, which will push it towards original equilibrium position, hence after releasing it will start oscillating about the original equilibrium or mean position

The situation has been shown in the figure

Now when the cork is at a distance X below mean position total height of cork inside liquid will be (X + Y) so volume of liquid displaced will be

V₂ = A(X+Y)

Here V₂ is the volume of liquid displaced and surface area of cork is A

So mass of liquid displaced by cork is

m₂ = ρ_l V₂

= ρ_l A(X+Y)

Where ρ_l is the density of liquid displacement

So weight of liquid displaced will be

W₂ = ρ_l A(X+Y)g

Where g is acceleration due to gravity

Now weight of liquid displaced will be the new force acting upward direction and, weight of the cylindrical cork is acting in downward direction so net force in upward direction is

F = W₂ – W

W is the weight of cork

We know W = W₁

Where W₁ is the weight liquid displaced by cork in equilibrium position

So we have

F = W₂ – W₁

Or

F = ρ_l A(X+Y)g - ρ_lAYg

= ρ_lAXg

We know force on a Body is

F = ma

Where m is the mass of particle and a is the acceleration of the particle, so acceleration of the body can be written as

a = F/m

here mass of the cork is

m = Ahρ

where A is the area of cross section of cork, ρ is its density and h is its height

so net acceleration of cork is

a = (ρ_lAXg)/(Ahρ)

= (ρ_lg/hρ)X

we know condition for simple harmonic motion is that acceleration of particle is proportional to distance from mean position and is directed toward mean position

and we have relation between acceleration and displacement as

a = -ω²X

where a is the acceleration of particle undergoing Simple Harmonic Motion with angular frequency ω

here acceleration of the cork is

a = -(ρ_lg/hρ)X

(-ve sign because acceleration is opposite to direction of displacement, acceleration is upward and displacement in vertically downward direction)

where the density of liquid ρ_l, the density of cork ρ, height of cork h, acceleration due to gravity g al are constant quantities so acceleration of cork is proportional to displacement from mean position or equilibrium position X i.e. Cork is undergoing simple harmonic motion, so comparing with the equation for acceleration of simple harmonic motion we get

ω² = ρ_lg/hρ

i.e. the angular frequency is

We know relation between time period T and angular frequency ω T = 2π/ω

So we have the time period of the oscillation as

So time period of the Simple harmonic motion of cork as

Now here the liquid on side of U-tube rises whereas on the other side of the U-tube containing mercury it has depressed by same amount due to the suction pump, the weight of the excess mercury will exert force in downward direction in downward direction to restore to original equilibrium position, let the area of cross section be A and ride in height of mercury be X

The situation has been shown in the figure

so total excess height of mercury on side of tube is 2x

now we know

Volume = Area × length

Here surface area of cross section of tube is A, excess height of mercury on side of tube is 2x so the excess volume of mercury on one side of the U-Tube is

V = 2xA

Now we know mass is given as

m = V × ρ

where m is the mass of the body, V is its volume and ρ is the density

so excess mass of mercury on side of tube is

m = 2Axρ

where A is the area of cross section of cork, ρ is density of mercury

we know weight of a body is given as

W = mg

Here m is the mass of the body and g is acceleration due to gravity, so weight of the excess mercury on one side of tube is

W = 2Axρg

Now this excess weight of mercury one side will exert force and tend to bring mercury in the tube to its original position hence the mercury in tube will start oscillating and height will decrease and increase periodically

now total mass of the mercury in U –Tube will be

M = Volume × Density

Let the total length of U – Tube be l and area of cross section is A we know

Volume = Area × length

so total Volume of liquid is

V_m = Al

So total mass of mercury is

M = V_mρ = Alρ

Where ρ is the density of mercury

We know force on a Body is

F = ma

Where m is the mass of Body and a is the acceleration of the Body, so acceleration of the body can be written as

a = F/m

here mass of the total mercury in the tube is

M = Alρ

where A is the area of cross section of cork, ρ is its density and l is the length of U - tube

so net acceleration of mercury in U-Tube is

a = -W/M

(negative sign because acceleration of mercury at each point inside tube is opposite to the direction of displacement)

i.e. a = -(2Axρg)/Alρ

= -(2g/l)x

we know condition for simple harmonic motion is that acceleration is proportional to distance from mean position and is directed toward mean position

and we have relation between acceleration and displacement as

a = -ω²x

where a is the acceleration of Body undergoing Simple Harmonic Motion with angular frequency ω, x is displacement from equilibrium or mean position

here in this case the acceleration of mercury is always directed towards mean position and also acceleration is given as

a = -(2g/l)x

here acceleration due to gravity g, length of tube l are constant so we can say acceleration is proportional to displacement from mean position, i.e. mercury in tube is undergoing simple harmonic motion, so comparing with the equation for acceleration of simple harmonic motion we get

ω² = 2g/l

i.e. the angular frequency is

We know relation between time period T and angular frequency ω T = 2π/ω

So we have the time period of the oscillation as

So time period of the Simple harmonic motion mercury in tube is

Here we are given av air chamber of volume V having a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction, suppose the ball is pushed down a bit, so the volume area of chamber decreases a bit, and hence pressure inside the chamber increases hence pressure on ball increases and pushes it in opposite direction, and since pressure variation is isothermal the product of pressure and volume will always be constant, also the force on ball will depend on its displacement from original position or mean position, so ball will start oscillating in an simple harmonic motion around the mean position, now let the ball be pushed downward by a distance x,

as shown in figure

Now we know

Volume = Area × length

So change in volume of air or decrease in volume will be

ΔV = a × X

Where a is the area of cross section of neck, X is the displacement of the ball or the length

Now force on the ball at any instant is

F = P × a

Where P is the pressure and a is the area of cross section

We know volumetric strain is given as

Strain = Change in Volume / Original Volume

Initially volume of chamber was V and it changed by ΔV

So the strain is

Strain = ΔV/V = aX/V

Now we know relation between stress and strain of air is given by bulk modulus of air as

B =Stress/Strain

We know stress is restoring force per unit area so here stress is the increase in pressure in the chamber due to decrease in volume of the chamber so the bulk modulus is

B = -P/(ΔV/V) = -P/(aX/V)

Negative sign suggest that volume has decreased when pressure has increased or the change is negative

So re-arranging above equation we get

B = -PV/aX

Or the pressure is

P = -BaX/V

We know

Now force on the ball at any instant is

F = P × a

Where P is the pressure and a is the area of cross section

So we get force on the ball at any instant as

F = -Ba²X/V

We know force on a particle is

F = mA

Where m is the mass of particle and a is the acceleration of the particle, so acceleration of a particle can be written as

A = F/m

now so we can say acceleration of the ball having mass m is

A = -Ba²X/mV

Or we can say

A = (-Ba²/mV)X

we know condition for simple harmonic motion is that acceleration of particle is proportional to distance from mean position and is directed toward mean position

and we have relation between acceleration and displacement as

A = -ω²X

where A is the acceleration of particle undergoing Simple Harmonic Motion with angular frequency ω

here in this case the force on ball is or acceleration of ball is always directed towards mean position and also acceleration is given as

A = -(Ba²/mV)X

here bulk modulus of air B, area of cross section of container a, mass of ball m, the original volume of container V all are constant so we can say acceleration of Ball is proportional to displacement from mean position, i.e. Ball is undergoing simple harmonic motion, so comparing with the equation for acceleration of simple harmonic motion we get

ω² = Ba²/mV

i.e. the angular frequency is

We know relation between time period T and angular frequency ω T = 2π/ω

So we have the time period of the oscillation as

So time period of the Simple harmonic motion of ball as

(a) Now there are four springs which support the entire weight of the car, assuming each member support equal weight,and the springs are identical under equilibrium condition the net upward force due spring and shock absorber system must be equal to weight of the weight of the car acting in downward direction

We know weight of a body is

W = Mg

Where M is the mass of the body and g is acceleration due to gravity

Here mass of car

M = 3000 kg

acceleration due to gravity

g = 10 ms^-2

so weight of the car is

W = 3000kg × 10ms^-2

= 3 × 10⁴N

Now all the four springs are at different positions along four wheels i.e. in parallel so if we assume spring constant of each spring as K so net equivalent spring constant will be

K_E = 4K

Now when a spring is compressed it exerts restoring force in opposite direction given as

F = Kx

Where F is the restoring force, K is the spring constant and x is the compression in the spring

Here the compression is

x = 15 cm = 0.15 m

the restoring force of spring due to compression balances the weight of the car so we have

F = W

K_Ex = Mg

4K × 0.15m = 3 × 10⁴N

Or we get spring constant of each member as

(b) Now since it is a system in which a mass is attached to spring so on slight disturbance from mean position, the body of car will undergo simple harmonic motion due to spring, no considering each spring, the mass supported by each spring is 1/4 th of the total mass i.e.

m = M/4 = 3000Kg/4 = 750 Kg

now the time period of oscillation of each spring undergoing Simple harmonic motion is given by relation

Where m is the mass attached to spring of spring constant K

Here the spring constant of each spring is

K = 5 × 10⁴ N/m

And mass attached to each spring is

m = 750 Kg

so we have time period of oscillations as

= 0.77 s

So time period of simple harmonic oscillation or un-damped oscillation of body of car is 0.77 s

But the motion is damped and amplitude of oscillation decreases by 50% during one complete oscillation or is reduced to half of original value after each complete oscillation

So for damping, to find amplitude at any instant we have the relation

A = A₀e^bt/2m

Where A₀ is the original amplitude of oscillation, A is the new amplitude after time t, of a particle of mass m undergoing damped oscillations with damping constant b

Here in this case If we let original amplitude be A₀ so new amplitude A will be 50% of A₀ or

A = A₀/2

Just After completion of one oscillation time is equal to time period i.e.

t = T = 0.77 s

mass attached to or supported by each spring

m = 750 Kg

so putting values we have

i.e.

taking natural Logarithm on both sides we get

(b×0.77s/1500 Kg) = ln2

b = (0.693 × 1500Kg)/0.77s

= 1350.28 Kg/s

So the damping constant b for the spring and shock absorber system of one wheel is 1350.28 Kg/s

Suppose a Body of mass m is undergoing Simple harmonic motion i.e. oscillating to and fro from a fixed position called mean Position with time period T of one oscillation, and the maximum displacement of the body from the mean position is called the amplitude A of the body, now the position of the particle or displacement of particle from mean position at any instant of time is given by the relation

x = A sin ωt

where x is denotes the position of particle or displacement from mean position at any instant of time t, ω is the angular frequency of the Simple harmonic motion of the block, A is the amplitude

differentiating the equation resenting position of particle with respect to time to find velocity of particle at any instant

We know where v is the velocity of the particle

And differentiation where a and b are constants

So differentiating we get velocity of the body as

v = Aω cos ωt

we know kinetic energy of a particle is given by the relation

K = 1/2 mv²

Where K is the kinetic energy of a particle of mass m and is moving with a velocity v

So instantaneous kinetic energy of the body undergoing Simple harmonic motion is

K = 1/2 m (A Aω cos ωt)²

Or Instantaneous kinetic energy of the body is

K = 1/2 mA²ω²cos²ωt

Now to find average kinetic Energy of the body in time interval T (time period of one complete oscillation) we will integrate the instantaneous kinetic energy with respect to time in the interval and then divide it with total time T

So average kinetic energy of the body denoted by K_avg will be

Since m,A, ω are constants we have

Solving further

We know relation between time period T and angular frequency ω ω = 2π/T

So we get

So the average kinetic energy of the body in one oscillation is

K_avg = 1/4 mA²ω²

Now the potential energy of a body undergoing Simple harmonic motion at any instant is given by the relation

U = 1/2 kx²

Where k is the force constant of the spring, and x is the displacement of the body from mean position

We know the relation between force constant and angular frequency as

k = m ω²

where k is the force constant, m is the mass of the body and v ω is the angular frequency of single harmonic motion

so the instantaneous potential energy of the body becomes

U = 1/2 mω²x²

And we know displacement of particle from mean position is

x = A sin ωt

where x is denotes the position of particle or displacement from mean position at any instant of time t, ω is the angular frequency of the Simple harmonic motion of the block, A is the amplitude

so putting the value of x we get, instantaneous Potential energy of the Body

U = 1/2 mA²ω²sin²ωt

Now to find average Potential Energy of the body in time interval T (time period of one complete oscillation) we will integrate the instantaneous Potential energy with respect to time in the interval and then divide it with total time T

So average Potential energy of the body denoted by U_avg will be

Since m,A, ω are constants we have

Solving further

We know relation between time period T and angular frequency ω ω = 2π/T

So we get

So the average Potential energy of the body in one oscillation is

U_avg = 1/4 mA²ω²

So we get that Average Potential Energy and Average Kinetic Energy of the Body undergoing Simple Harmonic Oscillation in one cycle or oscillation are equal and

K_avg = U_avg = 1/4 mA²ω²

Suppose a circular disc is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The restoring force due to wire will cause the disc to return to original position which depends on Torsional spring constant of wire α which is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist

The disc will start rotating to and fro around mean position and will undergo simple harmonic motion

The disc and its orientation has been shown in the figure

We are given

Period of torsional oscillations is T = 1.5 s

The radius of the disc is R = 15 cm=0.15 m

Mass of the disc is m = 10 Kg

Now Time period of body undergoing such torsional oscillations is given as

Where T is the time Period of oscillations, I is the moment of inertial of the body and α is the torsional spring constant of wire

Now we are given a disc and disc id rotating about an axis passing through its centre and perpendicular to the surface, moment of inertia of a circular disc in this orientation is given by the relation

I = 1/2 mR²

Where I is the moment of inertia of the disc, m is the mass of the disc of radius R

Here mass of the disc is

m = 10 Kg

radius of the disc is

R = 15 = 0.15 m

So moment of Inertia of the disc is

I = 1/2 × 10Kg × (0.15m)²

= 1/2 × 10Kg × 0.0225m²

= 0.1125 Kgm²

So the moment of inertia of the disc is

I = 0.1125 Kgm²

Now rewriting the equation

Squaring both sides we get

Or we get the torsional spring constant of the wire as

putting the values in above equation

i.e. the torsional spring constant of the wire is 1.97 Nm/rad

When a body describes simple harmonic motion it goes away from its original mean position up to a maximum distance called as the amplitude of the simple harmonic motion, then return towards the mean position and goes to the other extreme position and again then return to the mean position, the total time taken during this complete cycle is called the Time Period

Position of a Particle Undergoing simple harmonic motion with time Period T has been shown in the figure

Here we are given Amplitude of simple harmonic motion as

A = 5 cm = 0.05m

Time Period of simple harmonic motion as

T = 0.2 s

The angular frequency of simple harmonic motion is given as

ω = 2π/T

Where ω is the angular frequency of simple harmonic motion and T is the time period of simple harmonic motion

So here angular frequency is

ω = 2π/0.2s = 10π rad/s

now velocity of a particle undergoing simple harmonic motion is given by the relation

where v is the velocity of particle undergoing simple harmonic motion with amplitude A and angular frequency ω and is at a distance of x from mean position

Acceleration of a particle undergoing simple harmonic motion is given by the relation

a = -ω²x

where a is the acceleration of particle undergoing simple harmonic motion with angular frequency ω and is at a distance of x from mean position

(-ve sign indicates that acceleration is always directed towards the mean position and is in opposite direction to the displacement of particle)

(a) here displacement of particle from mean position is

x = 5 cm = 0.05 m

Here we are given Amplitude of simple harmonic motion as

A = 5 cm = 0.05m

here angular frequency is

ω = 10π rad/s

so putting values in the relation

we get

so velocity of the particle is 0 m/s

to find the acceleration we use the relation

a = -ω²x

so putting the values in the relation we get

a = -(10πs^-1)²×0.05m

= -100π²s^-2×0.05m

= -5π² ms^-2

So acceleration of the particle is 5π² ms^-2 directed towards the mean position

(b) here displacement of particle from mean position is

x = 3 cm = 0.03 m

Here we are given Amplitude of simple harmonic motion as

A = 5 cm = 0.05m

here angular frequency is

ω = 10π rad/s

so putting values in the relation

we get

so velocity of the particle is 0.4π m/s

to find the acceleration we use the relation

a = -ω²x

so putting the values in the relation we get

a = -(10πs^-1)²×0.03m

= -100π²s^-2×0.03m

= -3π² ms^-2

So acceleration of the particle is 3π² ms^-2 opposite to the velocity of the particle directed towards the mean position

(c) here displacement of particle from mean position is

x = 0 cm = 0 m

Here we are given Amplitude of simple harmonic motion as

A = 5 cm = 0.05m

here angular frequency is

ω = 10π rad/s

so putting values in the relation

we get

so velocity of the particle is 0.5π m/s

to find the acceleration we use the relation

a = -ω²x

so putting the values in the relation we get

a = -(10πs^-1)² × 0m

= -100π²s^-2 × 0m

= 0 ms^-2

So acceleration of the particle is 0 ms^-2

Suppose a block held on a frictionless plane attached to a spring(upstretched) is pulled to a distance x₀ and then pushed towards the centre that is the mean position or original position with velocity v_0, now the block will start to oscillate and undergo Simple Harmonic Motion with an angular frequency ω, the situation at beginning of time t = 0 has been

shown in the figure

Let the equation of Simple harmonic motion of the block be

x = a cos (ωt+θ)

where x is denotes the position of particle or displacement from mean position at any instant of time t, ω is the angular frequency of the Simple harmonic motion of the block, θ is the initial phase of the block a is the amplitude or the maximum displacement of block from the mean position

we have to find the amplitude a

differentiating the equation resenting position of particle with respect to time to find velocity of particle at any instant

We know where v is the velocity of the particle

And differentiation where a, b and c are constants

So differentiating we get velocity of particle

v = -aω sin (ωt+θ)

now we have initial condition that initially at t = 0, particle is at a distance x₀ from mean position

or x = x₀ at t = 0

so putting in equation

x = a cos (ωt+θ)

we have

x₀ = a cosθ (eq -1)

now we have initial condition that velocity of particle is initially is v₀ directed towards mean position in negative x direction

or v = v₀ at t = 0

so putting in equation

v = -aω sin (ωt+θ)

we get

v₀ = -aω sinθ

or we can rewrite it as

-v₀/ω = a sin θ (eq -2)

Squaring and adding (eq -1) and (eq -2) we get

x₀²+ (-v₀/ω)² = (a cos θ)² + (a sin θ)²

Solving further

x₀² + v₀²/ω² = a²cos²θ + a²sin²θ

or

a²(cos²θ + sin²θ) = x₀² + v₀²/ω²

we know the identity cos²θ + sin²θ = 1

so we have

a² = x₀² + v₀²/ω²

or the amplitude is

so the amplitude of oscillation of the block is