Oscillations

1. Answer: (b)and (c)
   (a) The swimmer’s motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.
   (b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time.
   (c) 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.
   (d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.
   
2. Answer: (b) and (c) are SHMs
   (a) and (d) are periodic, but not SHMs
   (a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.
   (b) An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.
   (c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.
   (d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.
3. Answer: (b)and (d) are periodic
   (a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.
   (b) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.
   (c) 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.
   (d) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.
4. (a) SHM
   The given function is:
   
   
   
   
   This function represents SHM as it can be written in the form:
   Its period is:
   (b) Periodic, but not SHM
   The given function is:
   
   
   The terms sin ωt and sin ωtindividually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.
   (c) SHM
   The given function is:
   
   
   This function represents simple harmonic motion because it can be written in the form:
   Its period is:
   (d) Periodic, but not SHM
   The given function is . 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 is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.
   (f) The given function 1 + ωt+ ω²t²is non-periodic.
5. Answer:
   (a) Zero, Positive, Positive
   (b) Zero, Negative, Negative
   (c) Negative, Zero, Zero
   (d) Negative, Negative, Negative
   (e) Zero, Positive, Positive
   (f) Negative, Negative, Negative
   Explanation:
   
   The given situation is shown in the following figure. 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
   (a) 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.
   (b) 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.
   (c)
   
   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.
   (d)
   
   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.
   (e)
   
   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.
   (f)
   
   This case is similar to the one given in (d).
    
6.  (c) A motion represents simple harmonic motion if it is governed by the force law:
   F= –kx
   ma= –k
   
   Where,
   Fis the force
   mis the mass (a constant for a body)
   xis the displacement
   ais the acceleration
   kis a constant
   Among the given equations, only equation a = –10 x is written in the above form with Hence, this relation represents SHM.
7. Initially, at t = 0:
   Displacement, x = 1 cm
   Initial velocity, v = ωcm/sec.
   Angular frequency, ω = πrad/s^–1
   It is given that:
   
   
   Squaring and adding equations (i) and (ii), we get:
   
   Dividing equation (ii) by equation (i), we get:
   
   SHM is given as:
   
   Putting the given values in this equation, we get:
   
   Velocity,
   Substituting the given values, we get:
   
   Squaring and adding equations (iii) and (iv), we get:
   
   Dividing equation (iii) by equation (iv), we get:
   
8. 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,
   g = acceleration due to gravity = 9.8 m/s²
   F = 50 × 9.8 = 490
   ∴Spring constant,
   Mass m, is suspended from the balance.
   Time period,
   
   ∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N
   Hence, the weight of the body is about 219 N.
9. Spring constant, k = 1200 N m^–1
   Mass, m = 3 kg
   Displacement, A = 2.0 cm = 0.02 cm
   (i) Frequency of oscillation v, is given by the relation:
   
   Where, T is the time period
   
   Hence, the frequency of oscillations is 3.18 cycles per second.
   (ii) Maximum acceleration (a) is given by the relation:
   a = ω² A
   Where,
   ω = Angular frequency =
   A = Maximum displacement
   
   Hence, the maximum acceleration of the mass is 8.0 m/s².
   (iii) Maximum velocity, v_max = Aω
   
   Hence, the maximum velocity of the mass is 0.4 m/s.
10. (a) x = 2sin 20t
    (b) x= 2cos 20t
    (c) x= –2cos 20t
    The functions have the same frequency and amplitude, but different initial phases.
    Distance travelled by the mass sideways, A = 2.0 cm
    Force constant of the spring, k = 1200 N m^–1
    Mass, m= 3 kg
    Angular frequency of oscillation:
    
    = 20 rad s^–1
    (a) When the mass is at the mean position, initial phase is 0.
    Displacement, x = Asin ωt
    = 2sin 20t
    (b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is.
    Displacement,
    
    = 2cos 20t
    (c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is.
    Displacement,
    = –2cos 20t
    The functions have the same frequencyand amplitude (2 cm), but different initial phases.
     
11. 
    (a) Time period, T = 2 s
    Amplitude, A = 3 cm
    At time, t = 0, the radius vector OP makes an angle with the positive x-axis, i.e., phase angle
    Therefore, the equation of simple harmonic motion for thex-projection of OP, at time t, is given by the displacement equation:
    
    
    
    (b) Time period, T = 4 s
    Amplitude, a = 2 m
    At time t = 0, OP makes an angle πwith the x-axis, in the anticlockwise direction. Hence, phase angle, Φ = + π
    Therefore, the equation of simple harmonic motion for thex-projection of OP, at time t, is given as:
    
    
12. (a)
    
    If this equation is compared with the standard SHM equation, then we get:
    
    The motion of the particle can be plotted as shown in the following figure.
    
    (b)
    If this equation is compared with the standard SHM equation, then we get:
    
    The motion of the particle can be plotted as shown in the following figure.
    
    (c)
    
    If this equation is compared with the standard SHM equation, then we get:
    Amplitude, A = 3 cm
    Phase angle, = 135°
    Angular velocity,
    The motion of the particle can be plotted as shown in the following figure.
    
    (d) x = 2 cos πt
    If this equation is compared with the standard SHM equation, then we get:
    Amplitude, A = 2 cm
    Phase angle, Φ = 0
    Angular velocity, ω = πrad/s
    The motion of the particle can be plotted as shown in the following figure.
    
13. (a) 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,
    For the two block system:
    The displacement (x) produced in this case is:
    
    Net force, F = +2 kx
    
    (b) For the one block system:
    For mass (m) of the block, force is written as:
    
    Where, x is the displacement of the block in time t
    
    It is negative because the direction of elastic force is opposite to the direction of displacement.
    
    Where,
    ω is angular frequency of the oscillation
    ∴Time period of the oscillation,
    
    For the two block system:
    
    
    It is negative because the direction of elastic force is opposite to the direction of displacement.
    
    Where,
    Angular frequency,
    ∴Time period,
14. Angular frequency of the piston, ω= 200 rad/ min.
    Stroke = 1.0 m
    Amplitude,
    The maximum speed (v_max) of the piston is give by the relation:
    
15. Acceleration due to gravity on the surface of moon, = 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
    
    Where,
    l is the length of the pendulum
    
    The length of the pendulum remains constant.
    On moon’s surface, time period,
    
    Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.
16. (a) The time period of a simple pendulum,
    For a simple pendulum, k is expressed in terms of mass m, as:
    km
    = Constant
    Hence, the time period T, of a simple pendulum is independent of the mass of the bob.
    (b) 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
    For large θ, sinθ is greater than θ.
    This decreases the effective value of g.
    Hence, the time period increases as:
    
    Where, l is the length of the simple pendulum
    (c) The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.
    (d) When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.
17. 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
    Where,
    vis the uniform speed of the car
    Ris the radius of the track
    Effective acceleration (a_eff) is given as:
    
    Time period,
    Where,l is the length of the pendulum
    ∴Time period, T
18. Base area of the cork =A
    Height of the cork = h
    Density of the liquid =
    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 g… (i)
    According to the force law:
    F = kx
    
    Where, k is a constant
    
    The time period of the oscillations of the cork:
    
    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:
    
19. 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
    Time period,
    Where,
    m is the mass of the mercury column
    Let l be the length of the total mercury in the U-tube.
    Mass of mercury, m= Volume of mercury × Density of mercury
    = Alρ
    ∴
    Hence, the mercury column executes simple harmonic motion with time period.
     
20. 
    Volume of the air chamber = V
    Area of cross-section of the neck = a
    Mass of the ball = m
    The pressure inside the chamber is equal to the atmospheric pressure.
    Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.
    Decrease in the volume of the air chamber, ΔV = ax
    Volumetric strain
    
    Bulk Modulus of air,
    In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.
    
    The restoring force acting on the ball,
    F =p × a
    
    In simple harmonic motion, the equation for restoring force is:
    F =–kx… (ii)
    Where, kis the spring constant
    Comparing equations (i) and (ii), we get:
    
    Time period,
    
    
21. (a) Mass of the automobile, m = 3000 kg
    Displacement in the suspension system, x = 15 cm = 0.15 m
    There are 4 springs in parallel to the support of the mass of the automobile.
    The equation for the restoring force for the system:
    F = –4kx = mg
    Where, k is the spring constant of the suspension system
    Time period,
    And = 5000 = 5 × 10⁴ N/m
    Spring constant, k = 5 × 10⁴ N/m
    (b) Each wheel supports a mass, M = = 750 kg
    For damping factor b, the equation for displacement is written as:
    
    The amplitude of oscillation decreases by 50%.
    ∴
    
    
    
    Where,
    Time period, = 0.7691 s
    = 1351.58 kg/s
    Therefore, the damping constant of the spring is 1351.58 kg/s.
22. The equation of displacement of a particle executing SHM at an instant tis given as:
    
    Where,
    A= Amplitude of oscillation
    ω= Angular frequency
    The velocity of the particle is:
    
    The kinetic energy of the particle is:
    
    The potential energy of the particle is:
    
    For time period T, the average kinetic energy over a single cycle is given as:
    
    
    And, average potential energy over one cycle is given as:
    
    It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.
     
23. 
    Mass of the circular disc, m = 10 kg
    Radius of the disc, r = 15 cm = 0.15 m
    The torsional oscillations of the disc has a time period, T = 1.5 s
    The moment of inertia of the disc is:
    I
    = × (10) × (0.15)²
    = 0.1125 kg m²
    Time period,
    αis the torsional constant.
    
    
    = 1.972 Nm/rad
    Hence, the torsional spring constant of the wire is 1.972 Nm rad^–1.
24. Amplitude,A= 5 cm = 0.05 m
    Time period, T= 0.2 s
    (a) For displacement, x= 5 cm = 0.05 m
    Acceleration is given by:
    a= –
    
    Velocity is given by:
    v
    
    = 0
    When the displacement of the body is 5 cm, its acceleration is
    –5π²m/s²and velocity is 0.
    (b) For displacement, x= 3 cm = 0.03 m
    Acceleration is given by:
    a=
    
    Velocity is given by:
    v
    
    = 0.4 πm/s
    When the displacement of the body is 3 cm, its acceleration is –3πm/s²and velocity is 0.4πm/s.
    (c) For displacement, x= 0
    Acceleration is given by:
    a= 0
    Velocity is given by:
    
    
    When the displacement of the body is 0, its acceleration is 0 and velocity is 0.5 πm/s.
25. The displacement equation for an oscillating mass is given by:
    x=
    Where,
    Ais the amplitude
    xis the displacement
    θis the phase constant
    Velocity,
    Att= 0, x= x₀
    x₀= Acosθ= x₀… (i)
    And,
    …(ii)
    Squaring and adding equations (i) and (ii), we get:
    
    Hence, the amplitude of the resulting oscillation is.