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Question-1

Which of the following examples represent periodic motion?

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

(ii) A freely suspended bar magnet displaced in its north-south direction and released

(iii) Halley's comet

(iv) A hydrogen molecule rotating about its centre of mass

(v) An arrow released from a bow.

Solution:
The cases (ii) and (iii) represents SHM as in both the cases the motion is to and fro about its mean position and the restoring force is proportional to displacement, but in opposite direction to the displacement.

The cases (i) and (iv) represent periodic motion as in case (i) there is no to and fro motion and in (iv) the polyatomic molecule has a number of natural frequencies and in general its vibration is a superposition of SHM's of a number of different frequencies. This superposition is periodic and not SHM.

Question-2

Which of the following examples represent SHM and which represent periodic but not SHM: 

(i) the rotation of the earth about its axis

(ii) the motion of an oscillating mercury column in a U-tube

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

(iv) the general vibration of a polyatomic molecule about its equilibrium configuration

Solution:
(ii) and (iii) represent SHM as in both cases the motion is to and fro about its mean position, and the restoring force is proportional to displacement, but is in opposite direction to the displacement.
(i) and (iv) represent periodic motion as in (i) there is no to and fro motion, and in (iv) the polyatomic molecule has a number of natural frequencies, and in general its vibration is a superposition of SHM's of a number of different frequencies. This superposition is periodic, and not SHM.

Question-3


 

 


Solution:
(a) It is not periodic as the motion is not repeated after a fixed point.

(b) It is periodic with time period of 2 seconds as x - t graph is repeated after 2 seconds.

(c) It is not periodic.

(d) It is periodic with a period of 2 seconds as the x - t graph is repeated after 2 seconds.

Question-4

Which of the following functions of time represent (a) SHM (b) periodic but not SHM and (c) non periodic motion? Give the period for each case of periodic motion (ω is any positive constant).

(i) sin
ωt - cos ωt

(ii) sin3
ωt

(iii) 3 cos (
π/4 -2ωt)

(iv) cos
ωt + cos 3ωt + cos 5ωt

(v)

(vi) 1 +
ωt + ω2t2?

Solution:
(i) f(t) = sin ωt - cos ωt ; Put 1 = r cos θ = r sinθ
    f(t) = r sin (ωt - θ)                             ---------(1)
    as r sin
θ= 1, r cos θ = 1
    r2 = (sin2
θ + cos2 θ) = 1 + 1 or r = 2
    From (1) we have, f(t) =
sin (ωt - θ)   ---------(2)
    Again, tan
θ = 1 or tan-1 (-1) = θ
    or tan-1 (tan π/4) = π/4

    Putting in (2) we get,
    f(t) =
2 sin (ωt - π/4) ---------(3)  
    Equation (3) can be written as, 
    f(t) =
2 sin
    Thus, f(t) represents SHM and its time period is given by,
                              T =

(ii) f(t) = sin3
ωt = 1/4 (3 sin 3 ωt - sin 3 ωt)
     It represents two SHMs separately, but their sum does not represent SHM.
     The period of function
3/4 sin ωt is given by T =, and the period of function 1/4 sin 3ωt is given by, 
     T
= =
     Thus the least time taken by the function to repeat the same value is T.
     T =
     Hence sin3
ωt is an example of a periodic function which is not SHM.

(iii) f(t) = 3cos
    
T = 
     Thus, 3 cos (2
ωt - π/4) represents SHM with, T =.

(iv) f(t) = cos
ωt + cos 3ωt + cos 5ωt
      The given function is a combination of the three independent SHMs, but their sum is not SHM.
      Time periods are = , .
      Therefore, the above function is periodic, but not SHM.

(v) f(t) = when t
, f(t) 0
    
It is a non-periodic function.

(vi) f(t) = 1 +
ωt + ω2t2 when t , f(t) .
      Thus, the above function is a non-periodic function.

Question-5

A particle is in linear SHM between two points A and B, 10cm apart. Take the directions from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(i) at the end A

(ii) at the end B

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

(iv) at 2cm away from B going towards A

(v) at 3cm away from A going towards B

(vi) at 4cm away from A going towards A

Solution:
(i) zero, +, +

(ii) zero, -, -

(iii) -, zero, zero

(iv) -, -, -

(v) +, +, +

(vi) -, +, + [here 'zero' stands for the magnitude of the quantity].

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 = 100x2.

Solution:
(c) a = -10x represents simple harmonic motion.

Question-7

The motion of a particle in SHM is described by the displacement function
x(t) = A cos (
ωt +φ), ω = 2π/T. 
If the initial position (t = 0) 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 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?

Solution:
x(t) = A cos (ωt + φ)
Initially t = 0, x =1, velocity =
π and angular frequency ω = π/s
A cos φ = 1                           ----------- (1)

Differentiating the given equation, we get
dx/dt = -A
ω sin (ωt + φ)
π = -Aω sin (ωt +φ

-A sin
φ = 1                             ----------- (2)

From equation (1) and (2) 
tan
φ = -1
φ = tan-1(-1)
= 3
π/4

Substituting the value of
φ in (1), we get
A cos (3
π/4) = 1
A =
2 cm
Thus the amplitude is found to be
2 cm and the initial phase is 3π/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 the period of 0.6 s. What is the weight of the body?

Solution:
When 50kg is put, the extension of spring is 20 cm. 
Force constant = k = Load / Extension
=
= 2450 Nm-1
Another body is suspended, its mass = m = ?
T = 0.6 s
T = 2
π
T2 = 4
π2
m =
= = 22.36 kg
Weight = mg = 22.36
× 9.8 = 219.36 N.

Question-9


 


Solution:
(i) Frequency of oscillation n = = = 3.18 Hertz

(ii) Maximum acceleration amax =
ω2a = a
                                    amax = 2
× 10-2 × 
                                    amax = 8 ms-2

(iii) Maximum velocity
νmax = aω 
                                      = 2
× 10-2 × 
                               νmax = 0.4 ms-1.

Question-10

In the previous problem, let us take the position of mass when the spring is unstretched at 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 compression position.   

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

Solution:
(a) We know that displacement at any instant t is expressed as y = a sin ωt
     If the displacement along x-axis is expressed by x, then for the given problem
     a = 2 cm
    
ω = 2πν = 2 ×3.14 ×3.2 = 20.1 s-1 
    
x = 2 sin (20.1 t)          ------(1)

(b) At the maximum stretched position, initial phase
ϕ = π/2.
    
x = a sin (ωt + ϕ) = a sin (ωt + π/2) = a cos wt
        x = 2 cos (20 t)                        ------(2)

(c) At the maximum compressed position,
θ = 3π/2
        x = a sin (
ωt + 3π/2) = a cos ωt
    or x = -2 cos (20 t)                       ------(3)

Therefore from the equations (1), (2) and (3), it is clear that the different functions for SHM do not differ in amplitude and frequency and varies only with the initial phase (
ϕ).

Question-11



 


Solution:
(a) x(t) = -3 sin πt, (b) x = -2 cos (π/2)t where x is in cm.

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 (2
πt + π/4)
(d) x = 2 cos
πt

Solution:
(a)                           
x = -2 sin
      = 2 cos
     = 2 cos ------(1)

Comparing equation (1) with the standard equation

x = A cos , we have
A = 2 cm

φ =
θ = 150°
= ω = 3 rad/s

(b)                         
x = cos ------(2)
Comparing equation (2) with the standard equation

x = A cos , we have
A = 1 cm

φ =
=
ω = 3 rad/s

(c)                         
x = 3 sin -------(3)
Equation (3) can be written as
x = -3 cos
  = -3 -------(4)
Comparing equation (3) with the standard equation

x = A cos , we have
A = 1 cm

φ =
=
ω = 1 rad/s

(d)                       
x = 2 cos
π t -------(5)
Comparing equation (5) with the standard equation
x = A cos , we have
A = 2 cm

φ = 0
ω = π

Question-13

 


Solution:
(i) The maximum extension x produced in the springs in figure (a) is given by constant   
 F = kx
 or        x = F/k
 The time period of oscillation is
 T =  

(ii) In case (a), one end A of the spring is fixed to the wall. When a force F is applied to the free end B in the direction shown in figure (a), the spring is stretched exerting a force on the wall which in turn exerts an equal and opposite reaction force on the spring, as a result of which every coil of the spring is elongated producing a total extension x. In case (b) shown in figure (b), both ends of the spring are free. Therefore, the reaction force is absent, as a result of which every coil of the spring is not elongated when force F is applied at each end in opposite directions. The coil at point O in the middle of the spring is not elongated. This situation can be visualized as two springs each of length l/2 (where l is the length of the complete springs) are joined to each other at point O. Since extension is proportional to the length of the spring, the force F applied at end B produces an extension x/2 in the part OB of the spring and the force F applied at A produces an extension x/2 in the part OA. The total extension in the spring is
Thus, the maximum extension produced in the spring in cases (a) and (b) is the same.
Now, the force constant of half spring is twice as that of the complete spring. In case (b) the force constant = 2k. Hence the time period of oscillation will be

T =

Question-14


 


Solution:
Case (i) Parallel combination of springs:
The figure (a) and (b) show the equilibrium state of the system. Suppose the mass is pulled downwards by a small distance x, the extension produced in each spring will be x. The restoring forces developed in the two springs are
F1 = -k1x and F2 = -k2x
Therefore total restoring force acting on the mass is
F = F1 + F2 = -k1x - k2x = -(k1 + k2)x = -kpx
where kp = k1 + k2 is the effective force constant of the parallel combination of springs. Now the time period of oscillation is given by

Tp =
Therefore frequency n =

Case (ii) Series combination of springs:
The figure (c) and (d) shows the equilibrium state of the system. Suppose the mass is pulled downwards by a small distance x. Let x1 and x2 be the extensions produced in springs of force constant k1 and k2 respectively. The restoring force developed in each spring will be the same say F.
Then        F = -k1x1 = -k2x2
so that x1 = F/k1 and x2 = -F/k2. Therefore, the total extension of the series combination is
x = x1 + x2
or           
F =
or           
F = -ksx

where ks = k1k2/(k1 + k2) is the effective force constant of the system. The time period of oscillation is given by

Ts =

Therefore frequency n =

Question-15


 


Solution:
We know T = 2π where m is the mass of the tray.
m =
When the tray is empty, t = 1.5 s, m = 12 kg, 

12 = -------(1)
k = = 1.1 x 102 N/m
When the block of mass m1 is added to the tray,
12 + m1 = --------(2)
From (1) and (2), = 4 or m1 = 36 kg.

Question-16

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 0.1 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rev/min, what is its maximum speed?

Solution:
Amplitude a = 0.1 m
Angular frequency w = 200 rev/min = 200
×2π/60 = 21 rad/s
Velocity v = aw cos wt = 0.1
×21 ×cos 41 ×1 = 1.01 m/s.

Question-17

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 surface of earth is 9.8 m/s2).

Solution:
Acceleration due to gravity on the surface of Moon gm = 1.7 ms-2
Acceleration due to gravity on the surface of Earth ge = 9.8 m/s2
Time period of the simple pendulum on the surface of Earth Te = = 3.5 s-1

Time period of the simple pendulum on the surface of Moon Tm

Tm =
=
= 8.4 s.

Question-18

Answer the following questions?
(i) Time period of the particle in SHM depends on the force constant k and mass m of the particle.
                                                
A simple pendulum executes SHM approximately. Why then is the time-period of a pendulum independent of the mass of the pendulum? 

(ii) The motion of the simple pendulum is approximately simple harmonic for small angles of oscillations. For larger angles of oscillations, a more involved analysis shows that T is greater than
         . Think of a qualitative argument to appreciate this result.

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

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

Solution:
(i) In case of a simple pendulum, force constant
k = (i.e. k is proportional to m)
T =
Thus, the time period is independent of the mass.

(ii) The force equation in the case of a simple pendulum,
F = -mg sin
θ = - mg θ (for small θ)
However, if
θ is not small then sin θ = θ, but sin θ < θ, which amounts to the reduction in the value of the g in the expression mg θ for large angles. Therefore, the time period for large amplitude is greater than 2π.  

(iii) Yes, because the wristwatch depends on the spring action and has nothing to do with gravity.

(iv) Gravity disappears for a man under free fall, so frequency is zero.

Question-19

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -at, where J is the restoring couple and t the angle of twist).

Solution:
Time period of torsional oscillations is given by
             T = 2
π                         -------(1)
where I is the moment of inertia of the disc about an axis coincident with the suspension wire and C is the torsional spring constant of the wire. Squaring the equation (i), we get
T2 =

Torsional spring constant, C =
I = 1/2 MR2
M = 10 kg
R = 15 cm = 0.15 m
T = 1.5 s

C =
      =
      = 2.0 nm/rad.

Question-20

A cylindrical piece of cork 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
                                            

where
ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Solution:
Area of cross-section of cork = A
Height of block = h
Density of the given liquid =
ρl
In equilibrium position the weight of the liquid displaced by the cork = weight of the cork.
When the cork is further depressed by a distance
ξ, the liquid displaces with an upthrust acting upwards that provides restoring force to the cork.
Hence Restoring force F = weight of the displaced liquid (upthrust)
                                   = -(Volume
×density ×g)
                                   = - A .
ξ . ρl . g
                            K = F/
ξ = -Aρl.g
The period of oscillation of cork is given by
                            T =
where m = mass of cork = volume of the cork
×density of cork = A.h.ρ
T =
  =

Question-21


 


Solution:
Mass of the trolley m = 3.0 kg
Force constant of each spring K = 600 Nm-1
Maximum displacement of the trolley from its mean position,
i.e., Amplitude, a = 5.0 cm
For the arrangement of springs shown in the figure, the resultant force constant
KR = K1 + K2 = K + K = 2K = 2
×600 Nm-1 = 1200 Nm-1

(a) The time period of oscillation of ensuring oscillations is given by
T = = 2
× = = 0.314 s

(b) Maximum velocity of the trolley, i.e.,
vmax = aw = a
×0.05 = 1 m/s

(c) Total energy of the trolley, i.e.,
E = = = 1.5 J.

Question-22

One end of a U-tube container 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.

Solution:
The restoring force,
Weight of mercury column of the height 2
ξ = F
F = -(volume)
×density ×g
F = -(A
×2ξ). πg = -2Aπgξwhere A = Area of cross-section of the tube  
ρ = density of mercury
K =  = 2A.
ρg
T =
Let L = length of the whole mercury column

Mass of mercury, m = Volume ×density = ALρ
T = =
where L is the total length of mercury column in the U-tube.

Therefore it shows that mercury column executes simple harmonic motion.

Question-23

An air chamber of volume V has a neck of area of cross-section a into which a ball of mass m can move without friction. Show that when the ball is pressed down some distance and released, the ball executes SHM. Obtain an expression for the time period of this SHM assuming pressure-volume variations to be (a) isothermal, (b) adiabatic?

Solution:
The ball of mass m is oscillating in the neck (area of cross-section A of an air chamber of volume V). When the ball is depressed by x, let the volume be reduced by v, and pressure increased by p.
(i) When the change is isothermal, we have,
 PV = (P + p) (V + v)
= PV - vP + pV - pv
or vP = pV (neglecting pv, which is very small)
or P =
= K (bulk modulus) \ P = K = -------(i)
         
or
Comparing it with F = Kx, we have = K, where K is force constant.

 
(ii) When the change is adiabatic,
           
(Using the binomial theorem neglecting higher terms)
        
      

Question-24

The bob of a vibrating simple pendulum is made of ice. How will the period of swing change when melting?

Solution:
The period of swing of simple pendulum will remain unchanged till the location of centre of gravity on the bob left after melting the ice remains at a fixed distance from the point of suspension. If the centre of gravity of ice bob after melting is raised upwards, then the effective length of pendulum decreases and hence the time period of swing decreases. If the centre of gravity of ice shifts on lower side, the time period of swing increases.

Question-25

The soldiers marching on a suspended bridge are advised to go out of step. Why?

Solution:
If the soldiers while crossing a suspended bridge march in step, the frequency of marching steps of soldiers may match with the natural frequency of oscillations of the suspended bridge. In that situation resonance will take place, then the amplitude of oscillations of the suspended bridge will increase enormously, which may cause the bridge to collapse. To avoid this situation, the soldiers are advised to go out of step on a suspended bridge.

Question-26

Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period?

Solution:
Average K.E. of a harmonic oscillator is given by
                    
 
     
Thus, we find that average kinetic energy and potential energy are equal.




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