Class 11 - Physics - Oscillations

Question1.

Which of the following examples represent periodic motion?

(a) A swimmer completing one (return) trip from one bank of a river to the other and back.

(b) A freely suspended bar magnet displaced from its N-S direction and released.

(c) A hydrogen molecule rotating about its centre of mass.

(d) An arrow released from a bow.

Answer:

  • It is not a periodic motion. Although the motion of the swimmer is to and fro but it does not have a definite time period.
  • The time taken by the swimmer during his to and fro journey may not the same.
  • It is a periodic motion. Once the freely suspended magnet is displaced and is allowed to execute motion, it will oscillate about its mean position with a definite time period.
  • It is a periodic motion. When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic.
  • It is not a periodic motion. As the arrow moves only in the forward direction. It does not come backwards. So this motion is not periodic.

Question2.

Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

(a) the rotation of earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

Answer:

  • It is periodic but not SHM. After rotation around its own axis earth comes to same position again and again in equal intervals of time.
  • As a result its motion is periodic. But as it does not execute to and fro motion about its own axis. Therefore it is not SHM.
  • It is a SHM. Mercury moves to and fro on the same path, about the fixed position, with a certain period of time.
  • It is simple harmonic motion. The ball is executing to and fro motion about the lowermost point in the bowl when released.
  • The ball is coming back to its initial position in the same period of time again and again. Therefore it executes SHM.

Question 3.

Figure 14.27 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion?

What is the period of motion (in case of periodic motion)?

Class_11_Gravitation_Graph1

Answer:

  • It is not periodic motion because the motion of the particle is not repeated.
  • It is a periodic motion as motion of the particle is repeated after a definite period of time. Time period is 2sec.
  • It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion,
  • the entire motion of the particle must be repeated in equal intervals of time.
  • It represents periodic motion with a time period of 2sec.

Question 4.

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic and (c) non-periodic motion?

Give period for each case of periodic motion (ω is any positive constant):

(a) (sin ωt – cos ωt)

(b) sin (3ωt)

(c) 3 cos (π/4 – 2ωt)

(d) (cos ωt + cos 3ωt + cos 5ωt)

(e) exp (–ω2t2)

(f) (1 + ωt + ω2t2)

 Answer:

(a) SHM                     (a) The given function is: (sin ωt – cos ωt)

This function represents SHM as it can be written in the form: a sin (ωt + Φ)

Its period is: (2π/ω)

(b) Periodic but not SHM

The given function is: sin (3ωt) = (1/4) [3 sin ωt - sin3ωt]

The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM).

However, the superposition of two SHM is periodic and not simple harmonic. Its period is: 2π/ω

(c) SHM

The given function is:

This function represents simple harmonic motion because it can be written in the form: a cos (ωt + Φ) its period is: (2π/2ω) = (π/ω)

(d) Periodic, but not SHM

The given function is (cosωt + cos3ωt + cos5ωt). Each individual cosine function represents SHM.

However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.

(e) Non-periodic motion

The given function exp (-ω2t2) is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

(f) The given function (1 + ωt + ω2t2) is non-periodic.

 Question 5.

A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart.

Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(a) at the end A,

(b) at the end B,

(c) at the mid-point of AB going towards A,

(d) at 2 cm away from B going towards A,

(e) at 3 cm away from A going towards B, and

(f) at 4 cm away from B going towards A.

Answer:

  • 0, (+) ive, (+) ive. The given situation is shown in the following figure.

Class_11_Gravitation_Figure4

Points A and B are the two end points, with AB = 10 cm. O is the midpoint of the path.

A particle is in linear simple harmonic motion between the end points

At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point.

Its acceleration is positive as it is directed along AO.

Force is also positive in this case as the particle is directed rightward.

  • 0, (-) ive, (-) ive. At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point.
  • Its acceleration is negative as it is directed along B. Force is also negative in this case as the particle is directed leftward.
  • (-) ive, 0, 0 .The particle is executing a simple harmonic motion. O is the mean position of the particle.
  • Its velocity at the mean position O is the maximum. The value for velocity is negative as the particle is directed leftward.
  • The acceleration and force of a particle executing SHM is zero at the mean position.

Class_11_Gravitation_Figure3 

  • (-) ive, (-) ive, (-) ive. The particle is moving toward point O from the end B.
  • This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle’s velocity and acceleration, and the force on it are all negative.

Class_11_Gravitation_Figure3

  • 0, (+) ive, (+) ive. The particle is moving toward point O from the end A.
  • This direction of motion is from A to B, which is the conventional positive direction. Hence, the values for velocity, acceleration, and force are all positive.

Class_11_Gravitation_Figure1

  • (-) ive, (-) ive, (-) ive. It is similar to the case (d).

Class_11_Gravitation_Figure2

Question 6.

Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

(a) a = 0.7x

(b) a = –200x2

(c) a = –10x

(d) a = 100x3

Answer:

Correct Option: - (c) a = -10x

In SHM, acceleration a is related to displacement by the relation of the form,

F =-kx, which is same as (c).

Question7.

The motion of a particle executing simple harmonic motion is described by the displacement function,

x(t) = A cos (ωt + φ ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle?

The angular frequency of the particle is π s–1.

If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α),

what are the amplitude and initial phase of the particle with the above initial conditions.

Answer:

Initially, at t = 0;

Displacement, x = 1 cm

Initial velocity, v = ω cm/ sec.

Angular frequency, ω = π rad/s–1

It is given that,

x (t) = A cos (ωt + Φ)

1 = A cos (ω × 0 + Φ) = A cos Φ

A cosΦ = 1                          ... (i)

Velocity, v= dx/dt

ω = -A ω sin (ωt + Φ)

1 = -A sin (ω × 0 + Φ) = -A sin Φ

A sin Φ = -1                         ... (ii)

Squaring and adding equations (i) and (ii), we get:

A2 (sin2 Φ + cos2 Φ) = 1 + 1

A2 = 2

∴ A = √2 cm

Dividing equation (ii) by equation (i), we get:

tan Φ = -1

∴Φ = 3π/4, 7π/4...

SHM is given as:

x = B sin (ωt + α)

Putting the given values in this equation, we get:

1 = B sin [ω × 0 + α] = 1 + 1

B sin α = 1                           ... (iii)

Velocity, v = ω B cos (ωt + α)

Substituting the given values, we get:

π = π B sin α

B sin α = 1                            ... (iv)

Squaring and adding equations (iii) and (iv), we get:

B2 [sin2 α + cos2 α] = (1 + 1)

B2 = 2   ∴ B = √2 cm

Dividing equation (iii) by equation (iv), we get:

B sin α / B cos α = 1/1

tan α = 1 = tan π/4

∴α = π/4, 5π/4...

 Question 8.

A spring balance has a scale that reads from 0 to 50 kg.

The length of the scale is 20 cm.

A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Answer:

Maximum mass that the scale can read, M = 50 kg

Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m

Time period, T = 0.6 s

Maximum force exerted on the spring, F = Mg

Where = acceleration due to gravity = 9.8 m/s2

F = 50 × 9.8 = 490 N

∴spring constant, k = (F/l) = (490/0.2) = 2450 N m-1.

Mass m, is suspended from the balance.

∴Weight of the body = mg = (22.36 × 9.8) = 219.167 N

Hence, the weight of the body is about 219 N.

Question 9.

A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 14.28.

A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Class_11_Gravitation_Spring1

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Answer:

Spring constant, k = 1200 N m–1

Mass, m = 3 kg

Displacement, A = 2.0 cm = 0.02 cm

Frequency of oscillation v is given by the relation:

v= (1/T) = (1/2π) √ (k/m)

Where, T = time period

Therefore, v = (1/ (2x3.14)) √ (1200/4)

=3.18m/s

Hence, the frequency of oscillations is 3.18 m/s.

Maximum acceleration (a) is given by the relation:

a = (ω2 A)

Where,

 ω = Angular frequency = √ (k/m)

A = Maximum displacement

Therefore, a = (k/m) A

= (1200 x 0.02)/ (3)

=8 ms-2

Hence, the maximum acceleration of the mass is 8.0 m/s2

Maximum velocity, vmax = (A ω)

=A√ (k/m)

=0.02 x√ (1200)/ (3)

=0.4m/s

Hence, the maximum velocity of the mass is 0.4 m/s.

Question 10.

In Exercise 14.9, let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of x-axis.

Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is 

 (a) at the mean position,

(b) at the maximum stretched position, and

(c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer:

a = 2cm, ω = (√ (k/m))

  • Since the time is measured from mean position, x =a sinωt = 2sin20t
  • At the maximum stretched position, the body is at the extreme right position. The initial phase = (π /2).

Therefore x =a sin (ωt + ((π /2)). = a cos ωt = 2 cos20t.

  • At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase = (3π /2).

Displacement, x =Asin (ωt + ((3π /2))

=2 sin (20t + ((3π /2)) = -2cos20t

The functions have the same frequency (20/2 π) Hz and amplitude (2cm) but with different phases (0, (π /2), (3π /2)).

Question 11.

Figures 14.29 correspond to two circular motions.

The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Class_11_Gravitation_Circular_Motion

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

Answer:

Given:

Time period, T = 2 s

Amplitude, A = 3 cm

At time t=0, the radius vector OP makes an angle (π/2) with the (+) ive x-axis, i.e. phase angle ϕ =+ (π/2)

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t, is given by the displacement equation:

x=A cos [(2 πt)/ (T) + ϕ]

=3 cos ((2 πt/2) + (π/2))

= -3sin (2 πt/2)

x= - 3sin πt cm

Time period, T = 4 s

Amplitude, a = 2 m

At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction. Hence, phase angle, Φ = (+ π)

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t, is given as:

x= a cos ((2 πt/T) + ϕ)

=2cos ((2 πt/4) + π)

Therefore, x = -2 cos (πt/2)m

Question 12.

Plot the corresponding reference circle for each of the following simple harmonic motions.

Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle.

For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).

(a) x = –2 sin (3t + π/3)

(b) x = cos (π/6 – t)

(c) x = 3 sin (2πt + π/4)

(d) x = 2 cos πt

 Answer:

  • x = -2 sin(3t + (π/3))

= + 2 cos (3t + (π/3) + (π/3))

=2 cos (3t + ((5π/6))

Comparing the above equation with the standard SHM equation

 x = A cos ((2 πt)/T + ϕ), then

Amplitude A = 2cm

Phase angle ϕ = (5π/6) = 1500

Angular velocity ω,   = (2π/T) = 3 rad/sec

The motion of the particle can be plotted as shown in the following figure.

 Class_11_Gravitation_Motion_Of_Particle1

x= cos ((π/6) – t) = cos (t – (π/6))

Comparing the above equation with the standard SHM equation

x = A cos ((2 πt)/T + ϕ),   then

Amplitude A = 1cm

Phase angle   ϕ= - (π/6) = -300

Angular velocity ω = (2π/T) = 1 rad/sec

The motion of the particle can be plotted as shown in the following figure.

x= 3 sin ((2π/T) + (π/4))

= -3 cos [(2πt + (π/4)) + (π/2)]

=-3cos (2πt + (3π/4))

Comparing the above equation with the standard SHM equation

x = A cos ((2 πt)/T + ϕ),   then

Amplitude, A = 3 cm

Phase angle, ϕ = (3 π /4) = 135°

Angular velocity, ω = (2π/T) = 2π rad/sec

The motion of the particle can be plotted as shown in the following figure.

Class_11_Gravitation_Motion_Of_Particle

Comparing the above equation with the standard SHM equation

x = A cos ((2 πt)/T + ϕ),   then

Amplitude, A = 2 cm

Phase angle, Φ = 0

Angular velocity, ω = π rad/s

The motion of the particle can be plotted as shown in the following figure

Question 13.

Figure 14.30 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end.

A force F applied at the free end stretches the spring.

Figure 14.30 (b) shows the same spring with both ends free and attached to a mass m at either end.

Each end of the spring in Fig. 14.30(b) is stretched by the same force F.

Class_11_Gravitation_Springs

(a) What is the maximum extension of the spring in the two cases?

(b) If the mass in Fig. (a) And the two masses in Fig. (b) Are released, what is the period of oscillation in each case?

Answer:

  • For the one block system:

When a force F, is applied to the free end of the spring, an extension l, is produced. For the maximum extension, it can be written as:

F = kl

Where, k is the spring constant

Hence, the maximum extension produced in the spring, l= (F/k)

  • For the two block system:

The displacement (x) produced in this case is:

x = (l/2)

Net force, F = +2 kx = 2k (l/2)

Therefore l = (F/k)

For the one block system:

For mass (m) of the block, force is written as:

F = ma = m (d2x/dt2)

Where, x is the displacement of the block in time t

Therefore, m (d2x/dt2) = -kx

It is negative because the direction of elastic force is opposite to the direction of displacement.

(d2x/dt2) = - (k/m) x = - ω2x

Where ω2 = (k/m)

Or ω = √ (k/m)

Where, ω = angular frequency of the oscillation

∴Time period of the oscillation, T = (2 π/ ω)

= (2 π)/ (√ (k/m)) = (2 π)/ (√ (m/k))

For the two block system:

It is negative because the direction of elastic force is opposite to the direction of displacement.

(d2x/dt2) = - [2k/m] x = -ω2x

Where, angular frequency ω = (√ (2k/m))
Therefore, Time period= T = (2 π)/ (ω)

= (2 π)/ (√ (m/2k))

Question 14.

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m.

If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

Answer:

Angular frequency of the piston, ω = 200 rad/ min.

Stroke = 1.0 m

Amplitude, A = (1.0/ (2)

= 0.5m

The maximum speed (vmax) of the piston is given by the relation:

vmax = A ω

= (200 x 0.5)

vmax =100 m/min

Question 15.

The acceleration due to gravity on the surface of moon is 1.7 m s–2.

What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s–2)

Answer:

Acceleration due to gravity on the surface of moon, g' = 1.7 m s–2

Acceleration due to gravity on the surface of earth, g = 9.8 m s–2

Time period of a simple pendulum on earth, T = 3.5 s

T=2π√ (l/g)

Where l =length of the pendulum

l= (T2)/ ((2π) 2x g)

= (3.5)2/ (4x (3.14)2) x 9.8 m

The length of the pendulum remains constant,

On moon’s surface, time period, T’= 2π√ (l/g’)

=2π ((3.5)2/4x (3.14)2 x 9.8)/ (1.7)

 =8.4s

Question 16.

Answer the following questions:

(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:

T = 2π √ (m/k). A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations.

For larger angles of oscillation, a more involved analysis shows that T is greater than 2π √ (l/g). Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

Answer:

  • For a simple pendulum, force constant or spring factor k is proportional to mass m; therefore, m cancels out in denominator as well as in numerator.
  • That is why the time period of simple pendulum is independent of the mass of the bob.

 In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:

F = –mg sinθ

Where, F = Restoring force, m = Mass of the bob,

g = Acceleration due to gravity, θ = Angle of displacement

For small θ, sinθ ≈ θ

For large θ, sinθ is greater than θ.

This decreases the effective value of g.

Hence, the time period increases as: T = 2π√ (l/g)

  • Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.
  • Gravity disappears for a man under free fall, so frequency is zero

Question 17.

A simple pendulum of length l and having a bob of mass M is suspended in a car.

The car is moving on a circular track of radius R with a uniform speed v.

If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Answer:

The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

Acceleration due to gravity = g

Centripetal acceleration = (v2/R)

Where, v = uniform speed of the car

R = radius of the track

Effective acceleration (aeff) will be:

aeff = √((g)2 + (v2/R)2)

Time period, T = 2π√ (l/ aeff)     

Where, l is the length of the pendulum

Therefore,

Time period, T = 2π√ (l)/ (√ ((g) 2 + (v2/R) 2))

 

Question 18.

A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl.

The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

T= 2π √ (hρl) / (gρl) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Answer:

Base area of the cork = A

Height of the cork = h

Density of the liquid = ρ1

Density of the cork = ρ

In equilibrium:

Weight of the cork = Weight of the liquid displaced by the floating cork

Let the cork be depressed slightly by x.

As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust = Restoring force, F = Weight of the extra water displaced

F = – (Volume × Density × g)

Volume = (Area × Distance through which the cork is depressed)

Volume = Ax

∴ F = – (A x ρ1 g)..... (i)

According to the force law:

F = kx

k = (F/x) where, k is constant

k = (F/x)

 = - (Aρ1 g).... (ii)

The time period of the oscillations of the cork:

T = 2π√ (m/k)      .... (iii)

Where,

m = Mass of the cork

= (Volume of the cork × Density)

= (Base area of the cork × Height of the cork × Density of the cork) = (Ahρ)

Hence, the expression for the time period becomes:

T=2π√ (Ahρ)/ (Ahρ1g)

T = 2π√ (hρ)/ (ρ1g)

Question 19.

One end of U-tube containing mercury is connected to a suction pump and the other end to atmosphere.

A small pressure difference is maintained between the two columns.

Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Answer:

Area of cross-section of the U-tube = A

Density of the mercury column = ρ

Acceleration due to gravity = g

Restoring force, F = Weight of the mercury column of a certain height

F = – (Volume × Density × g)

F = – (A × 2h × ρ ×g)

= (–2Aρgh)

= (–k × Displacement in one of the arms (h))

Where,

2h is the height of the mercury column in the two arms

k is a constant, given by k= -(F/h) = 2Aρg

Time period,

T = 2π√ ((m/k)

= 2π√ ((m)/ (2Aρg)

Where, m = the mass of the mercury column

Let l = length of the total mercury in the U-tube.

Mass of mercury, m = (Volume of mercury × Density of mercury)

= Alρ

T=2π√ (Alρ)/ (2Aρg)

=2π √ ((l/2g)

Hence, the mercury column executes simple harmonic motion with time period

T = 2π √ ((l/2g)

 

 

 

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