Solution-Class-12-Physics-Chapter-5-Oscillations-MSBSHSE

(A) \(\frac{\sqrt{3}}{2}A\)

(D) 18 J

(A) 1/6 m     

(B) 1:2    

(B) The force is maximum at time 3T/4.

It is defined as the linear periodic motion of a body, in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position. OR

A particle is said to execute linear SHM if the particle undergoes oscillations about a point of stable equilibrium, subject to a linear restoring force always directed towards that point and whose magnitude is proportional to the magnitude of the displacement of the particle from that point.

• Examples : The vibrations of the tines (prongs) of a tuning fork, the oscillations of the needle of a sewing machine.

The general expression for the displacement of a particle in SHM at time t is

x = A sin(ωt + α) ….. (1)

where A and ω is a constant in a particular case and α is the initial phase.

The velocity of the particle is 

\(v=\frac{dx}{dt}=\frac{d}{dt}[A sin(ωt + α)]\)

= ωA cos(ωt + α)

= \(ωA\sqrt{1-sin^2(ωt + α)}\)

From Eq. (1), sin(ωt + α) = x/A

\(v=ωA\sqrt{1-\frac{x^2}{A^2}}\)

∴ \(v=ω\sqrt{A^2-x^2}\)  …… (2)

Equation (2) gives the velocity as a function of x

The acceleration of the particle is

a= \(\frac{dv}{dt}=\frac{d}{dt}\)[Aω cos(ωt + α)]

∴ a = − ω²A sin(ωt + α)

But from Eq. (1),  A sin(ωt + α) = x

∴ a = − ω²x  ……. (3)

Equation (3) gives the acceleration as a function of x. The minus sign shows that the direction of the acceleration is opposite to that of the displacement.

Consider a simple pendulum of length L, suspended from a rigid support O. When displaced from its initial position of rest through a small angle q in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in Fig.

At B, the forces on the bob are its weight mg and the tension T’  in the string. Resolve mg into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.

mg cos θ is balanced by the tension in the string.

mg sin θ restores the bob to the equilibrium position.

Restoring force, F = − mg sin θ

If θ is small and expressed in radian,

sin θ  » θ = \(\frac{arc}{radius}=\frac{AB}{OB} =\frac{x}{L}\)

∴ F = − mgθ = − mg\(\frac{x}{L}\)     ….(1)

Since m, g and L are constant,

∴ F ∝ (−x)  …….(2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.

Acceleration, a = F/m =  \(\frac{g}{L}x\)  ......(3)

Therefore, acceleration per unit displacement

\(\left | \frac{a}{x} \right |=\frac{g}{L} \)

Period of SHM,

T = \(\frac{2π}{\sqrt{\text{acceleration per unit displacement}}}=\frac{2π}{\sqrt{g/L}}\)

∴ T = \(2π\sqrt{L/g}\)

This gives the expression for the period of a simple pendulum.

The period of a simple pendulum at a given place is

T = \(2π\sqrt{L/g}\)

where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows

Law of length :The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length. T ∝ \(\sqrt{L}\)

Law of mass :The period of a simple pendulum does not depend on the mass or material of the bob of the pendulum.

Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity. T ∝ \(\sqrt{\frac{1}{g}}\)

Law of isochronism :The period of a simple pendulum does not depend on the amplitude of oscillations, provided that the amplitude is small.

Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth's magnetic field is B. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North—South direction, along B.

If it is rotated in the horizontal plane by a small displacement θ from its rest position (θ =0),the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.

The bar magnet experiences a torque r due to the field B. Which tends to restore it to its original orientation parallel to B. For small θ, this restoring torque is

τ = −μB sin θ = −μBθ ...(1)

where the minus sign indicates that the torque is opposite in direction to the angular displacement θ.

Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional in magnitude of the angular displacement but opposite in direction. Hence, for small angular displacment, the oscillations of the bar magnet in a uniform magnetic field is simple harmonic.

Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth's magnetic field is B. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North—South direction, along B.

If it is rotated in the horizontal plane by a small displacement θ from its rest position (θ =0),the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.

The bar magnet experiences a torque r due to the field B. Which tends to restore it to its original orientation parallel to B. For small θ, this restoring torque is

τ = −μB sin θ = −μBθ ...(1)

where the minus sign indicates that the torque is opposite in direction to the angular displacement θ.

For clockwise angular displacement θ , the restoring torque is in the anticlockwise direction.

∴ τ = Iα = −μBθ

where I is the moment of inertia of the bar magnet and a is its angular acceleration.

∴ α = −μBθ/I --- (1)

Since μ, B and I are constants, Eq. (1) shows that angular acceleration is directly proportional to the angular displacement and directed opposite to the angular displacement.

Hence the magnet performs angular S.H.M. The period of vibrations of the magnet is given by

∴ T =  \(\frac{2π}{\sqrt{α/θ}}=\frac{2π}{\sqrt{\text{angular acceleration per unit angular displacement}}}\)

∴ T = \(\frac{2π}{\sqrt{α/θ}} = 2π\sqrt{\frac{I}{−μB}\)

Consider a particle which moves anticlockwise around a circular path of radius r with a constant angular speed w (Fig).

Let the path lie in the x-y plane with the centre at the origin O. The instantaneous position P of the particle is called the reference point and the circle in which the particle moves as the reference circle.

The perpendicular projection of P onto the y-axis is Q. Then, as the particle travels around the circle, Q moves to-and-fro along the y-axis. Line OP makes an angle a with the x-axis at t= 0. At time t, this angle becomes θ = ωt + α.

The projection Q of the reference point is described by the y-coordinate,

y = OQ = OP sin ∠OPQ, Since  ∠OPQ = ωt + α.

y = r sin(ωt + α.)

which is the equation of a linear SHM of amplitude A. The angular frequency w of a linear SHM can thus be understood as the angular velocity of the reference particle.

Projection of velocity: The tangential velocity of the reference particle is

v = ωr. Its y-component at time t is‘

v_y = ωr sin (90°− θ) = ωr cos θ

v_y = ωr cos (ωt + α)

Projection of acceleration: The centripetal acceleration of the reference particle is a_r = ω²r , so that its y-component at time t is

a_x = a_r sin ∠OPQ

∴ a_x = − ω²r sin (ωt + α) = − ω²y

(a) Conclusions from the graphs for the mean position:

• Displacement, velocity and acceleration of S.H.M. are periodic functions of time.
• Displacement time curve and acceleration time curves are sine curves and velocity time curve is a cosine curve.
• There is phase difference of π/2 radian between displacement and velocity.
• There is phase difference of π/2 radian between velocity and acceleration.
• There is phase difference of π radian between displacement and acceleration.
• Shapes of all the curves get repeated after 2π radian or after a time T.
• The phase differences are independent of the initial phase

(v) Conclusions for positive extreme position :

• The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting from an extreme position, the x—t and a—t graphs are cosine curves; the v—t graph is sine curve.
• There is a phase difference of π/2 radians between x and v, and between v and a.
• There is a phase difference of π radians between x and a.
• The phase differences are independent of the initial phase

Graph :

Consider a particle of mass m performing linear SHM with amplitude A. The restoring force acting on the particle is P = −kx, where k is the force constant and x is the displacement of the particle from its mean position.

Kinetic energy : At distance x from the mean position, the velocity is

 v = \(ω\sqrt{A^2−x^2}\)

where ω  = \sqrt{\frac{k}{m}}\) . The kinetic energy (KE) of the particle is

KE = ½ mv²

= ½ mω²(A² − x²)

= ½ k(A² − x²)  ……..(1)

If the phase of the particle at an instant t is 0 = ωt + α, where a is initial phase, its velocity at that instant is

v = ωA cos (ωt + α)

and its KE at that instant is

KE = ½ mv² = ½ mω²A² cos²(ωt + α)

=  ½ kA² cos²(ωt + α)  …..(2)

Therefore, the KE varies with time as cos²θ .

Potential energy : The potential energy of a particle in linear SHM is defined as the work done by an external agent, against the restoring force, in taking the particle from its mean position to a given point in the path, keeping the particle in equilibrium.

Suppose the particle in fig. is displaced from P₁ to P₂, through an infinitesimal distance dx against the restoring force F as shown.

The corresponding work done by the external agent will be dW = (- F )dx = kx dx. This work done is stored in the form of potential energy.

The potential energy (PE) of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.

∴ PE = \(\int dW =\int kx\,dx =\frac{1}{2}dx^2 \)

The displacement of the particle at an instant t being P

x = A sin (ωt + α)

its PE at that instant is

PE = ½ kx² = ½ kA² sin (ωt + α)  …….(4)

Therefore, the PE varies with time as sin²θ.

Total energy : The total energy of the particle is equal to the sum of its potential energy and kinetic energy.

From Eqs. (1) and (2), total energy is

E = PE + KE

= ½ kx² + ½ k(A² – x²)  

= ½ kx² + ½ kA² – ½ k x² 

= ½ kA² = ½ mω²A² = …….(5) 

Total energy of SHM in terms of acceleration is

E = ½ mω²A² = ½ mAa_max         (….a_max = ω²A)

∴ E =  ½ mAa_max  is the the expression of total energy in terms of acceleration.

As m is constant, and ω and A are constants of the motion, the total energy of the particle remains constant (or its conserved).

Factors on which the total energy of a particle executing SHM depends :

If η is the frequency in S.H.M. ω = 2πη and for T is the period of the motion w = 2π/T

Using this in Eq. (5), we get

= 2π²mη²A² = \(\frac{2π^2mA^2}{T^2}\)

Conclusions :The total energy of the particle is

• independent of x and thus remains constant when m, w and A are constant,
• directly proportional to the force constant (E ∝ k),
• directly proportional to the mass of the particle (E ∝ m),
• directly proportional to the square of the amplitude (E ∝ A²),
• proportional to the square of the frequency η (E ∝ η² ),
• inversely proportional to the square of the period T (E ∝ 1/T²).

Consider a simple pendulum of length L, suspended from a rigid support O. When displaced from its initial position of rest through a small angle q in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in Fig.

At B, the forces on the bob are its weight mg and the tension T’  in the string. Resolve mg into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.

mg cos θ is balanced by the tension in the string.

mg sin θ restores the bob to the equilibrium position.

Restoring force, F = − mg sin θ

If θ is small and expressed in radian,

sin θ  » θ = \(\frac{arc}{radius}=\frac{AB}{OB} =\frac{x}{L}\)

∴ F = − mgθ = − mg\(\frac{x}{L}\)     ….(1)

Since m, g and L are constant,

∴ F ∝ (−x)  …….(2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.

Acceleration, a = F/m =  \(\frac{g}{L}x\)  ......(3)

Therefore, acceleration per unit displacement

\(\left | \frac{a}{x} \right |=\frac{g}{L} \)

Period of SHM,

T = \(\frac{2π}{\sqrt{\text{acceleration per unit displacement}}}=\frac{2π}{\sqrt{g/L}}\)

∴ T = \(2π\sqrt{L/g}\)

This gives the expression for the period of a simple pendulum.

From the above expression, the factors on which its period depends.

• The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length. T ∝ \(\sqrt{L}\)
• The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity. T ∝ \(\sqrt{\frac{1}{g}}\)

Given : v =  \(\frac{1}{2}v_max\) , 2A=10 cm

 A=5 cm

v = \(ω\sqrt{A^2-x^2}\)   and v_max = ωA

Since v = \(\frac{1}{2}v_max\)

\(ω\sqrt{A^2-x^2}=\frac{ωA}{2}\)

∴ \(A^2-x^2=\frac{A^2}{4}\)

∴ \(x^2=A^2-\frac{A^2}{4} = \frac{3A^2}{4} \)

∴ \(x= ±\frac{\sqrt{3}}{2}A\)

= 0.866 x 5= ± 4.33 cm

This gives the required displacement.

Given :  \(\frac{d^2x}{dt^2}=-36x\)

Comparing this equation with the general equation,

We get,  ω² = 36,  ∴ ω = 6 rad/s

ω = 2πη

The frequency, η = ω/2π = 6/(2x3.142) =  0.9548 Hz

and the period, T = 1/η = 1/0.9548 = 1.047s

Given : A=4 cm = 4 x 10^-2m, η = 5Hz, t= 1/30 s

ω = 2πη = 2π x 5 = 10π rad/s

Therefore, the magnitude of the acceleration,

|a| = ω²x = ω²A sin ωt         

= (10π)² (4 X 10^-2) sin 10π/30

= 4π² sin π/3 = 4 (9.872)(0.866) = 34.20 m/s²

Given : PE = 0.1π²x² J, m = 20 g = 2 X 10^-2 kg

PE = ½ mw²x² = ½m(4π²η²)x² = 0.1π²x²

∴ 2mη² = 0.1

∴ η² =  0.1/[2(2 X 10^-2)]= 2.5

∴ The frequency of SHM is

η = \(\sqrt{2.5}\) = 1.581 Hz

Given : m = 2 kg, E = 40 J

The speed of the body while crossing the centre of the path (mean position) is v_max and the total energy is entirely kinetic energy.

  \(\frac{1}{2}\)mv²_max = E

∴  v_max = \(\sqrt{\frac{2E}{m}}= \sqrt{\frac{2×40}{2}}\) = 6.324 m/s

Given : T = 4 s, x = A/3

The displacement of a particle starting into SHM from the mean position is

x =A sin ωt = A sin(2π/T)t

∴ A sin(2π/T)t = A/3

∴ (2π/T)t = sin^-1 0.3333 = 19.47° = 19.47 x π/180  rad

∴ t = 19.47/90 = 0.2163s

∴ The displacement of the bob will be one-third of its amplitude 0.2163s after crossing the mean position.

Given : L = 100cm, m = 50 g = 5 x 10⁻² kg, x = 3 cm, g = 9.8 m/s²

Restoring force, F = mg sin θ = mgθ

= (5 x 10⁻²)(9.8)(3/100)

= 1.48 x 10⁻²N

Given : g₁ =9.75 m/s², g₂ = 9.8 m/s²

Length of a seconds pendulum, L = g/π²

∴ L₁ = g₁/π² = 9.75/9.872 = 0.9876 m and

∴ L₂ = g₂/π² = 9.8/9.872  = 0.9927 m

 The length of the seconds pendulum must be increased from 0.9876 m to 0.9927 m, i.e., (0.9927−0.9876) by 0.0051 m.

Given : A=8 cm, KE = 3PE

KE = \(\frac{1}{2}\)k(A² —x²) and PE = \(\frac{1}{2}\)kx²

Given, KE = 3PE.

∴ \(\frac{1}{2}\) k(A² —x²) = 3( \(\frac{1}{2}\) kx²)

∴ A² —x² = 3x² ,  ∴ 4x² = A²

the required displacement is

x = ± A/2 = ±8/2 = ±4 cm

Given : T = 2π s, v = \(b\sqrt{3}\)   m/s at x=b

ω = 2π/T = 2π/2π =1 rad/s

v = ω\(\sqrt{A^2-x^2}\)

∴ at x = b, \(b\sqrt{3}\)  = (1)\(\sqrt{A^2-x^2}\)

∴ 3b² = A² – b² ,   ∴ A = 2b

∴ Assuming the particle starts from the mean position, its displacement is given by

x = A sin ωt = 2b sint

If the particle is at x = b at t = t₁,

b = 2b sin t₁  ∴ t₁= sin^-1(1/2) = π/6 s

Also, with period T = 2π s, on travelling a further distance b the particle will reach the positive extremity at time t₂ = π/2 s.

 The time taken to travel a further distance b from  x = b is t₂ – t₁ = π/2 − π/6 = π/3 s

For mass m₁, period is T

∴ T = 2π\(\sqrt{\frac{m_1}{k}}\)

For mass m₂, period is 2T

∴ 2T = 2π\(\sqrt{\frac{m_1+m_2}{k}}\)

∴ 2T/T = 2 = \(\sqrt{\frac{m_1+m_2}{m_1}}\)

∴ \(\frac{m_1+m_2}{m_1}\) = 4

∴ \(\frac{m_2}{m_1}\) = 3

∴ \(\frac{m_1}{m_2}\) = \(\frac{1}{3}\)

x= a sin ωt + b cos ωt

Let a = A cos ∅ and b = A sin ∅, so that

A² = a² + b² and tan ∅ = b/a

 x=A cos ∅ sin ωt + A sin ∅ cos ωt

 x = A sin (ωt+∅)

which is the equation of a linear SHM with amplitude  A = \(\sqrt{a^2+b^2}\)   and phase constant ∅ = tan^-1(b/a) as required.

Given : x₁ = 5sin (4πt + π/3) = A₁ sin(ωt + α),

x₂ = 3sin (4πt + π/4) = A₂ sin(ωt + β),

∴ A₁ =5 cm, A₂ =3 cm, α = π/3 rad, β = π/4 rad

(i) Resultant amplitude,

R = \(\sqrt{A_1^2+A_2^2+2A_1A_2cos(α-β)}\)

∴ R = \(\sqrt{5^2+3^2+2(5)(3)cos(π/3-π/4)}\)

= \(\sqrt{25+9+30cos(π/3-π/4)} = \sqrt{62.98}\)

= 7.936 cm

(ii) Epoch of the resultant SHM,

δ = \(tan^{-1}(\frac{A_1sinα+A_2sinβ}{A_1cosα+A_2cosβ})\)

= \(tan^{-1}(\frac{5sin(π/3)+3sin(π/4)}{5cos(π/3)+3cos(π/4)})\)

= \(tan^{-1}(\frac{5(0.866)+3(0.7071)}{5(0.5)+3(0.7071)})\)

= \(tan^{-1}(1.396)\) = 54⁰23’

Given: R = 10cm = 0.1 m, M = 0.2 kg, θ_m=60⁰ = p/3 rad, T=1 s, π³ = 31

The MI of the disc about the rotation axis (perpendicular through its centre) is

I = ½ MR² = ½ (0.2)(0.1)² = 10^-3 kg-m²

The period of torsional oscillation, T = 2π\(\sqrt{\frac{I}{c}}\)

∴ The torsion constant, c = 4π²\(\frac{I}{T^2}\)

The magnitude of the maximum restoring torque,

τ_max = cθ_m = 4π²\(\frac{I}{T^2}(\frac{π}{3})\) 

= \(\frac{4}{3}π^3\frac{I}{T^2}=\frac{4}{3}(31)(\frac{10^{-3}}{1^2})\) 

= 41.33 x 10^-3 = 0.04133 N-m

Given:  B =1.6 x 10^-5 T, I = 3 x 10^-6 Kg/m², m =3 A-m²

The period of oscillation, T = 2π\(\sqrt{\frac{I}{μB}}\)

∴ The frequency of oscillation is

η = \(\frac{1}{2π}\sqrt{\frac{μB}{I}}\)

The number of oscillations per minute

60η =\(\frac{60}{2π}\sqrt{\frac{(3)(1.6×10^{-5})}{3×10^{-6}}}\)

=\(\frac{60}{2π}\sqrt{16}\)

= 38.19 per minute

Given : A = 0.25m, g = π² = 10 m s²

During vertical oscillations, the acceleration is maximum at the turning points at the top and bottom. The block will just lose contact with the piston when its apparent weight is zero at the top,

i.e., when its acceleration is a_max = g, downwards.

|a_max| = ω²A = 4π² η²_max A = g

∴ η²_max  = \(\sqrt{\frac{g}{4π^2A}}\) = \(\sqrt{\frac{10}{4×10×0.25}}\) = 1 Hz

This gives the required frequency of the piston.

E = ½ mω²A² = ½ m(4π² η²)A²

∴ E/m = 2p² h²A² = 2 x 10 x 1² x (1/4)² = 1.25 J/kg