Physics-16.9-Oscillations
SIMPLE HARMONIC MOTION AND OSCILLATION
Simple Harmonic motion (SHM):
Any motion which repeats itself after regular interval of time is called periodic or harmonic motion.
If a particle in periodic motion moves back and forth (or to and fro) over the same path, then its motion is called oscillatory or vibratory. The examples of oscillatory or vibratory motion are:
1. the motion of a pendulum
2. the motion of a spring fixed at one end, which is stretched or compressed and then released
3. the motion of a violin string
4. the motion of atoms in molecules or in a solid lattice
5. the motion of air molecules as a sound wave passes by
Conditions of Simple Harmonic Motion
For SHM is to occur, three conditions must be satisfied.
1. There must be a position of stable equilibrium
At the stable equilibrium potential energy is minimum.
That is, and
2. There must be no dissipation of energy
3. The acceleration is proportional to the displacement and opposite in direction.
That is,
1.1 TYPES OF MOTION
(i) PERIODIC MOTION
When a body or a moving particle repeats its motion along a definite path after regular intervals of time, its motion is said to be Periodic Motion and interval of time is called time or harmonic motion period (T). The path of periodic motion may be linear, circular, elliptical or any other curve.
(ii) OSCILLATORY MOTION
'To and Fro' type of motion is called an Oscillatory Motion. It need not be periodic and need not have fixed extreme positions.
The force/torque acting in oscillatory motion (directed towards equilibrium point) is called restoring force/torque.
(iii) EQUATION OF SIMPLE HARMONIC MOTION (SHM):
The necessary and sufficient for a SHM = Force constant
F = –kx
where k = Force constant for a SHM,
x = displacement from mean position.
or
It's solution is
1.2 CHARACTERISTICS OF SHM
(i) Amplitude: It is the maximum value of displacement of the particle from its equilibrium position.
(ii) Time period (T): Smallest time interval after which the oscillatory motion gets repeated is called Time period.
(iii) Frequency (f): Number of oscillations competed in unit time interval is called frequency of oscillations, its units is or Hz.
(iv) Angular Frequency (w): The quantity is called the angular frequency of the oscillating system. As we know that second order differential equation of simple harmonic motion is called angular frequency and its units is rad/sec.
(v) Phase: The physical quantity which represents the state of motion of particle (eg. its position and direction of motion (orientation) at any instant).
In the solution of sedond order differential equation of SHM, is called phase of the motion.
(vi) Displacement (x)
If time is measured from the equilibrium position , displacement x from equilibrium point at any instant of time t is given by x = Asint
(vii) Velocity (v): v = (A sint) = A cost
or v = A =
(a) Velocity is minimum at extreme positions and is zero.
At x = A, v = vmin = zero.
(b) Velocity is maximum at the equilibrium position and is A.
At x = 0, v = vmax = A
(c) Direction of velocity is either towards or away from the equilibrium position.
(viii) Acceleration (a) a = = -2A sint = -2x
(a) The minimum value of acceleration is zero and it occurs at equilibrium.
(b) The maximum value of acceleration is 2A and it occurs at extreme positions.
(c) Acceleration is always directed towards the equilibrium position and so it is always opposite to the direction of the displacement.
1.3 ENERGY OF A BODY IN S.H.M.
(i) Potential Energy
The linear restoring force acting on the harmonic oscillator is given by
F = m = kx
Now if the oscillator is displaced through a further displacement dx against the force, work done in displacing the particle is given by
dW = k dx
Hence the total work done in displacing the particle from mean position (x=0) to (x=x) is given by
W =
By convention P.E. at the mean position is taken as zero. Hence, above equation gives the values of P.E. of harmonic oscillator at a displacement x from the mean position i.e.,
U = . . . (1)
This shows the P.E. is proportional to the square of the displacement and graph showing the variation of potential energy with the displacement will be a parabola given by continuous line in the figure. P.E. is maximum at maximum displacement and is given by
Umax =
(ii) Kinetic Energy
Velocity of harmonic oscillator is given by equation as
v = =
Hence kinetic energy of the oscillator is given by
K.E. = . . . (2)
The graph showing the variation of K.E. with x is shown in figure by dotted line.
The kinetic energy is maximum when x = 0. Thus
K.E.max =
Now total energy E of the oscillator for displacement x is given by
E = P.E. + K.E. =
= = constant . . . (3)
Thus total energy is independent of the displacement, it remains constant throughout the motion of the oscillator. Also the total energy is equal to maximum value of either P.E. or K.E.
(iii) Average Value of P.E. and K.E.
By equation (1) P.E. at distance x is given by
U =
The average value of P.E. for one complete oscillation is given by
dt
= =
Because the average value of sine or of cosine function for the complete cycle is equal to zero.
Now K.E. at x is given by
=
The average value of K.E. for one complete cycle
Taverage =
= =
Thus average values of K.E. and P.E. of harmonic oscillator are equal and each is equal to half of the total energy.
Exmaple 1: A particle executes S.H.M. with time period 4s find the time taken by the particle to go directly from its mean position to half its amplitude.
Solution:
At
Hence,
or
or
as
SIMPLE HARMONIC MOTION IN SPRING-MASS SYSTEM
Let us find out the time period of a spring-mass system oscillating on a smooth horizontal surface as shown in the figure.
At the equilibrium position the spring is relaxed. When the block is displaced through a distance x towards right, it experiences a net restoring force F = –kx towards left.
The negative sign shows that the restoring force is always opposite to the displacement. That is, when x is positive, F is negative, the force is directed to the left. When x is negative, F is positive, the force is directed to the right. Thus, the force always tends to restore the block to its equilibrium position x = 0.
F = -kx
Applying Newton's Second Law,
or
Comparing the above equation with, we get
or
Exmaple 2: Find the period of oscillation of a vertical spring-mass system.
Solution: Let be the deformation in the spring in equilibrium.
Then
When the block is further displaced by x, the net restoring force is given by
or
Using second law of motion
or
Thus,
Exmaple 3: A block of mass m attached to a spring of stiffness k is placed on smooth horizontal surface, as shown in the figure. The other end of the spring is attached to a rigid vertical wall. When the spring is in relaxed position, the net force on the block is zero. If the block is slightly displaced from its equilibrium position spring force acting on the body has tendency to bring it at the equilibrium.
Solution: Equation of motion of the body is given by
Where x = elongation in the spring .
Hence the body will execute S.H.M. Comparing with the equation,
we get, as T 2/
Exmaple 4: For the arrangement shown in the figure, find the period of oscillation.
Solution: Obviously, when the block displaced down by x, the spring will stretch by .
From the free body diagram of the pulley,
or
The net restoring force on the block is T. Using the Second Law of motion, we get
or
Thus, the period of SHM is given by
Series and Parallel Combinations of Springs
When two springs are joined in series, the equivalent stiffness of the combination may obtained as
When two springs are joined in parallel, the equivalent stiffness of the combination is given by
Fig. (a) Series combination of spring (b) Parallel combination of spring
Exmaple 5: A spring of stiffness constant k and natural length is cut into two parts of length and respectively, and an arrangement is made as shown in the figure. If the mass is slightly displaced, find the time period of oscillation.
Solution: The stiffness of a spring is inversely proportional to its length. Therefore the stiffness of each part is
and
Time period,
or
Exmaple 6: Two masses and are connected by a spring of force constant k and are placed on a frictionless horizontal surface. Show that if the masses are displaced slightly in opposite directions and released, the system will execute simple harmonic motion. Calculate the frequency of oscillation.
Solution: Let masses and be displaced by and respectively from their equilibrium position in opposite direction so that the total extension in the spring will be . Due to this stretch a restoring force kx will act on each mass and so equation of mass will be
i.e. … (i)
while that for m2 will be
i. e. … (ii)
But as x x1 x2 i.e. … (iii)
So substituting equation (i) and (ii) in (iii),
or
or
Exmaple 7: For the arrangement shown in figure, the spring is initially compressed by 3cm. When the spring is released the block collides with the wall and rebounds to compress the spring again.
(a) If the coefficient of restitution is 0.7, find the maximum compression in the spring after collision.
(b) The spring is released at t = 0. Find the minimum time after which the block becomes stationary.
Solution: (a) Velocity of the block just before collision,
or v0 =
here, x0 = 0.03 m, x = 0.01 m, k = 104N/m, m = 1 kg
v0 =
After collision , v = ev0 = (0.7) = 2 m/s.
Maximum compression in the spring is
or xm¬ = =2.23 cm
(b) In the case of spring – mass system, since the time period is independent of the amplitude of oscillation,
therefore,
T =
T =
SUPERPOSITION OF TWO SHM'S
(a) In same direction and of same frequency.
If , both SHM's are in phase and
If , both SHM's are out of phase and
The resultant amplitude due to superposition of two or more than two SHM's of this case can also be found by phasor diagram also.
(b) In same direction but are of different frequencies. (special case, if )
then resultant displacement
(c) In two perpendicular directions
Case(i) if So path will be straight line & resultant displacement will be
Case (ii) if
so, resultant will be . i.e. equation of an ellipse and if A = B, then superposition will be an equation of circle.
TYPES OF PENDULUM
4.1 COMPOUND PENDULUM
When a rigid body is suspended from an axis and made to oscillate about that then it is called compound pendulum.
C = Initial position of center of mass
C' = Position of center of mass after time t
S = Point of suspension
= Distance between point of suspension and center of mass (it remains constant during motion)
where = moment of inertia relative to the axis which passes from the center of mass & parallel to the axis of oscillation.
where
k = gyration radius (axis passes from centre of mass)
= equivalent length of simple pendulum;
T is minimum when
4.2 TORSIONAL PENDULUM
Consider a body, such as a disc or a rod, suspended at the end of a wire, as shown in the fig.
Fig. Torsional Pendulum
When the end of the wire is twisted by an angle , the restoring torque is given by
where k is called the torsional constant.
Using Newton's law
or
Thus, the angular frequency and the time period of oscillation are given by
and
4.3 PHYSICAL PENDULUM
Any rigid body suspended from a fixed support constitutes a physical pendulum. Consider the situation when the body is displaced through a small angle Torque on the body about o is given by
(1)
where = distance between point of suspension and centre of mass of the body.
If I be the M.I. of the body about O.
then (2)
From (1) and (2), we get
as and are oppositely directed.
since is very small.
Comparing with the equation , we get
Exmaple 8: A rod of mass M and length L is pivoted about its end O as shown in the figure(10.19). Find the period of SHM.
Fig. A rod suspended about its end acts like a physical pendulum
Solution: Restoring torque is
Using Newton's Second Law and the small angle approximation, we get
Using parallel-axes theorem
Thus,
MOTION OF A LIQUID IN U-SHAPED TUBE
m : mass of the liquid in the tube
: density of the liquid
A : cross sectional area of the tube
If the tube is of U-shape and liquid is filled to a height h,
then 1 = 2 = 90° and m = hAd 2
Exmaple 9: An ideal liquid of density occupies length of an U – tube of uniform cross-section as shown in the figure(10.25). When the liquid surface is displaced through x the liquid column executes SHM. Find the period of oscillation.
Solution: When the liquid column is displaced through x on the right side, the weight of the unbalanced liquid provides the restoring force.
where A is the area of cross-section Applying Newton's second law
or
Time period is
SHM OF A PARTIALLY SUBMERGED FLOATING OBJECT
If it is connected to a vertical spring of spring constant k
SOLVED OBJECTIVE PROBLEMS
1. A flat horizontal board moves up and down in SHM of amplitude. Then the shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is
(A) (B)
(C) (D)
Solution:(B)
2. A uniform spring of force constant k is cut into two pieces, the lengths of which are in the ratio 1 : 2. The ratio of the force constant of the shorter and the longer pieces is
(A) 1 : 2 (B) 2 : 3
(C) 2 : 1 (D) 1 : 3
Solution:(C)
3. The displacement of a particle moving in SHM at any instant is given by y = sin t. The acceleration after time is (where T is the time period)
(A) 2 (B) -2
(C) - (D)
Solution:(B) When , the particle is at the extreme position. At the extreme position, the particle has maximum acceleration -2.
4. A simple pendulum is suspended from the roof a trolley which moves in a horizontal direction with an acceleration . Then the time period is given by T = 2 , where g’ is equal to
(A) g (B) g -
(C) g + (D)
Solution:(D)
5. A pendulum suspended from the ceiling of a train has a period T when the train is at rest. When the train is accelerating with a uniform acceleration, the period of oscillation will
(A) increase (B) decrease
(C) remain unaffected (D) become infinite
Solution:(B) Comparing with
Clearly, T’ mg cos 20
(C) < mg cos 20 (D) mg cos 40
Solution:(B) In this case, T- mg cos 20 = centripetal force
T> mg cos 20
11. Two masses m1 and m2 are suspended together by a massless spring of spring constant K. When the masses are in equilibrium, m1 is removed without disturbing the system. The angular frequency of m2 is
(A) (B)
(C) (D)
Solution:(B) (m1+m2)g = K(l1 + l2) …(i)
After the removal of mass m1,
m2g = Kl2 …(ii)
(i) – (ii) gives m1g = Kl1 or l1=
This gives the amplitude of oscillation of mass m2.
Now, T = 2
12. In Q.11, The amplitude of oscillation of m2 is
(A) (B)
(C) (D)
Solution:(A) (m1+m2)g = K(l1 + l2) …(i)
After the removal of mass m1,
m2g = Kl2 …(ii)
(i) – (ii) gives m1g = Kl1 or l1=
This gives the amplitude of oscillation of mass m2.
Now, T = 2
13. A block of mass 0.1 kg is held between two rigid supports by two springs of force constants 8 N m-1 and 2 N m-1. If the block is displaced along the direction of the length of the springs, then the frequency of vibration is
(A) (B)
(C) (D)
Solution:(A) Springs are connected in parallel.
Combined spring constant,
14. Two identical springs, each of spring constant K, are connected in series and parallel as shown in fig. 4. A mass m is suspended form them. The ratio of their frequencies of vertical oscillations will be
(A) 2 : 1 (B) 1 : 1
(C) 1 : 2 (D) 4 : 1
Solution:(C)
15. Refer to Fig. 5. One kg block performs vertical harmonic oscillations with amplitude 1.6 cm and frequency 25 rad s-1. The maximum value of the force that the system exerts on the surface is
(A) 20 N (B) 30 N
(C) 40 N (D) 60 N
Solution:(D) Weight = 5.1 9.8 N = 49.98 N = 50 N
Maximum restoring force
Maximum force on the surface = (50 + 10) N = 60 N.
16. A small spherical heavy ball of radius r is placed on a smooth concave mirror of radius of curvature R placed on a horizontal table. If the sphere is displaced slightly form the position of rest, then it executes SHM with a period of
(A) (B)
(C) (D)
Solution:(C) The effective radius is R – r. For calculation of time period, refer to important terms, facts and formulae.
17. A pendulum clock that keeps correct time on the Earth is taken to the Moon. It will run
(A) at correct rate (B) 6 times faster
(C) times faster (D) times slower.
Solution:(D) g decreases by a factor of 6. T increases by a factor of . So, the clock is times slower.
18. A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 2.5 cm. What must be the least period of these oscillation so that the object is not detached ? (Given : g = 10 m s-2)
(A) s (B)
(C) (D)
Solution:(C)
19. A forced oscillator is acted upon by a force F = F0 sin t. The amplitude of the oscillator is given by . What is the resonant angular frequency ?
(A) 2 units (B) 9 units
(C) 18 units (D) 36 units
Solution:(B) At resonance, amplitude should be maximum.
This is possible if 22 - 36 + 9 is minimum.
20. Two simple harmonic motions are represented by :
. The ratio of the amplitudes of two SHM is
(A) 1 : 1 (B) 1 : 2
(C) 2 : 1 (D) 1 :
Solution:(A) In the second case, amplitude is
i.e.
So, the ratio of amplitudes is 1 : 1.
21. The vertical extension in a light spring by a weight of 1 kg is 9.8 cm. The period of oscillation is
(A) (B)
(C) (D)
Solution:(A)
22. The period of a simple pendulum, whose bob is a hollow metallic sphere, is T. The period is T1 when the bob is filled with sand, T2 when it is filled with mercury and T3 when it is half filled with sand. Which of the following is true?
(A) T = T1 = T2>T2 (B) T1 = T2 > T3>T
(C) T > T3 > T1=T2 (D) T = T1 = T2T, the clock will lose the time.
=
Time lost in t = 1 day is
=
= s = 2.7 s
38. A pendulum has time period T for small oscillations. An obstacle P is situated below the point of suspension O at a distance The pendulum is released from rest. Throughout the motion the moving string makes small angle with vertical. Time after which the pendulum returns back to its initial position is
(A) T (B)
(C) (D)
Solution: (B) After P length of pendulum becomes
Now as T , so after P time period will become T’ = T/2, Therefore, the desired time will be :
t =
39. The two block of mass m1 and m2 are kept on a smooth horizontal table as shown in figure. Block of mass m1 but not m2 is fastened to the spring. If now both the blocks are pushed to the left so that the spring is compressed a distance d. The amplitude of oscillation of block of mass m1, after the system is released is
(A) (B)
(C) (D)
Solution: (A) Block of mass m2 shoots off carrying some kinetic energy away from the system. To find its speed. Potential energy of spring = maximum kinetic energy of blocks.
(k = force constant of spring)
or v2 =
with m1 alone on the spring :
maximum potential energy = maximum kinetic energy of m1.
or
or KA2 =
or A =
40. A particle starts SHM from the mean position. It’s amplitude is a and total energy E. At one instant its kinetic energy is its displacement at this instant is
(A) y = (B) y =
(C) y = (D) y = a
Solution: y = A sin t
E =
K.E. = =
A2 y2 = A2
y =
41. A particle executing SHM while moving from one extremity is found at distance x1, x2, and x3, from the centre at the end of three successive seconds. The time period of oscillation is
(A) (B)
(C) (D)
Solution: (B) Displacement time equation of the particle will be
x = A cos
Given that x1 = A cos
x2 = A cos 2
and x3 = A cos 3
Now =
=
or T = where
42. Two masses M and m are suspended together by a massless spring of force constant k. When the masses are in equilibrium, M is removed without disturbing the system. The amplitude of oscillation is
(A) (B)
(C) (D)
Solution: (D) For equilibrium of (M + m)
X1 =
and for equilibrium of m
x2 =
amplitude of oscillation will be
A = x1 – x2 =
43. A block is kept on a rough horizontal plank. The coefficient of friction between the block and the plank is 1/2. The plank is undergoing SHM of angular frequency 10 rad/s. The maximum amplitude of plank in which the block does not slip over the plank is : (g = 10 m/s2)
(A) 4cm (B) 5cm
(C) 10 cm (D) 16 cm
Solution: (B) Maximum acceleration in SHM is
amax = =
this will be provided to the block by friction.
Hence amax = or =
or A = =
= 0.05 m = 5 cm
44. A particle executes SHM between x = -A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then
(A) T1 < T2 (B) T1 > T2
(C) T1 = T2 (D) T1 = 2T2
Solution: Let x = A sin t
From 0 – A/2
A/2 = A sin T1
or sin T1 = = sin
or T1 =
From 0 – A
A = A sin T’1
or, sin T’1 =1 = sin ( / 2)
or, or T’1 =
From A/2 – A :
T2 = T’1 T1 = =
T1 = and T2 =
T1 < T2
45. A simple pendulum of length 1m with a bob of mass m swings with an angular amplitude 30o. Then tension in the string at angular displacement 15o: (g = 9.8 m/s2)
(A) greater than mg sin15oC
(B) greater than mg cos15oC
(C) greater than mg cos75oC
(D) greater than mg cot75oC
46. A simple pendulum of length 1m with a bob of mass m swings with an angular amplitude 30o. Then rate of change of speed at angular displacement 15o: (g = 9.8 m/s2)
(A) g cos 15o
(B) g cot 25o
(C) g sin 15o
(D) g sin 75o
Solution: (45-46) (B,C) Simple pendulum of length 1 m is called second’s pendulum whose time period is T = 2s. But this time period is for small oscillations. In this case angular amplitude is 30o. Therefore, time period will not be 2s.
At angular displacement 15o.
T – mg cos15o =
Or T > mg cos15o.
And tangential acceleration at = g sin 15o = rate of change of speed.
47. A constant force F is applied on a spring block system as shown in figure. The mass of the block is m and spring constant is k. The block is placed over a smooth surface. Initially the spring was unstretched. Then motion is which execute type?
(A) SHM (B) Circular
(C) straight line (D) rotation
48. A constant force F is applied on a spring block system as shown in figure. The mass of the block is m and spring constant is k. The block is placed over a smooth surface. Initially the spring was unstretched. Then what is time period of oscillation?
(A) (B)
(C) (D)
49. A constant force F is applied on a spring block system as shown in figure. The mass of the block is m and spring constant is k. The block is placed over a smooth surface. Initially the spring was unstretched. Then what is the maximum speed of block?
(A) (B)
(C) (D)
Solution: (47-49)(A,C,D) The situation is similar as if a block of mass m is suspended from a vertical spring and a constant force mg acts downwards. Therefore, in this case also block will execute SHM with time period:
T =
At compression x where
F = kx
X =
This is also the amplitude of oscillation.
Hence
A =
At mean position speed of the block will be maximum. Applying work energy theorem
F. x =
or v =
50: If a SHM is represented by the equation in S.I. Units Determine its amplitude?
(A) 10 m (B) 1 m
(C) 6 m (D) 14 m
51: If a SHM is represented by the equation in S.I. Units Determine its time period?
(A) 6 s (B) 2 s
(C) 10 s (D) 3 s
52: If a SHM is represented by the equation in S.I. Units Determine its maximum velocity Q?
(A) (B)
(C) (D)
Solution 1:(50-52)(A, B, C) Comparing the above equation with
we get,
53. The spring mass system is shown in the figure. The spring stretches 2 cm from its free length when a force of 10 N is applied. This spring is stretched 10 cm from its free length, a body of mass m = 2kg attached to it and released from rest at time t = 0. Find the force constant of the spring.
Write the equation of motion of the body in the from x = A sin ( + ) where x is the displacement from the equilibrium position. Express the spring force as a function of time.
(A) 500 N/m (B) 800 N/m
(C) 300 N/m (D) 100 N/m
54. The spring mass system is shown in the figure. The spring stretches 2 cm from its free length when a force of 10 N is applied. This spring is stretched 10 cm from its free length, a body of mass m = 2kg attached to it and released from rest at time t = 0. Find the time period
Write the equation of motion of the body in the from x = A sin ( + ) where x is the displacement from the equilibrium position. Express the spring force as a function of time.
(A) 0.307 s (B) 0.397 s
(C) 0.297 s (D) 5.300 s
55. The spring mass system is shown in the figure. The spring stretches 2 cm from its free length when a force of 10 N is applied. This spring is stretched 10 cm from its free length, a body of mass m = 2kg attached to it and released from rest at time t = 0. Find the frequency of vibration
Write the equation of motion of the body in the from x = A sin ( + ) where x is the displacement from the equilibrium position. Express the spring force as a function of time.
(A) 7.20 Hz (B) 5.51 Hz
(C) 2.51 Hz (D) 0.55 Hz
56. The spring mass system is shown in the figure. The spring stretches 2 cm from its free length when a force of 10 N is applied. This spring is stretched 10 cm from its free length, a body of mass m = 2kg attached to it and released from rest at time t = 0. Find the amplitude of vibration.
Write the equation of motion of the body in the from x = A sin ( + ) where x is the displacement from the equilibrium position. Express the spring force as a function of time.
(A) 3.06 m (B) 1.00 m
(C) 0.06 m (D) 8.03 m
57. The spring mass system is shown in the figure. The spring stretches 2 cm from its free length when a force of 10 N is applied. This spring is stretched 10 cm from its free length, a body of mass m = 2kg attached to it and released from rest at time t = 0. Find the initial velocity and acceleration
Write the equation of motion of the body in the from x = A sin ( + ) where x is the displacement from the equilibrium position. Express the spring force as a function of time.
(A) 1, 1 m/s2 upward (B) 2, 7 m/s2 upward
(C) 3, 10 m/s2 upward (D) 0, 15 m/s2 upward
58. The spring mass system is shown in the figure. The spring stretches 2 cm from its free length when a force of 10 N is applied. This spring is stretched 10 cm from its free length, a body of mass m = 2kg attached to it and released from rest at time t = 0. Find the maximum velocity and acceleration
Write the equation of motion of the body in the from x = A sin ( + ) where x is the displacement from the equilibrium position. Express the spring force as a function of time.
(A) 0.95 m/s, 15 m /s2 (B) 0.95 m/s , 25 m /s2
(C) 0.95 m/s , 5 m /s2 (D) 0.95 m/s , 2 m /s2
59. The spring mass system is shown in the figure. The spring stretches 2 cm from its free length when a force of 10 N is applied. This spring is stretched 10 cm from its free length, a body of mass m = 2kg attached to it and released from rest at time t = 0. Find the spring force at the two extreme position of the body.
Write the equation of motion of the body in the from x = A sin ( + ) where x is the displacement from the equilibrium position. Express the spring force as a function of time.
(A) x = 00.06 sin (-t + /2) (B) x = 00.06 sin (t + /2)
(C) x = 00.06 sin (t - /2) (D) x = 00.01 sin (t + /2)
60. The spring mass system is shown in the figure. The spring stretches 2 cm from its free length when a force of 10 N is applied. This spring is stretched 10 cm from its free length, a body of mass m = 2kg attached to it and released from rest at time t = 0. Find the time taken by the body to move half way towards the equilibrium position from its initial position.
Write the equation of motion of the body in the from x = A sin ( + ) where x is the displacement from the equilibrium position. Express the spring force as a function of time.
(A) 20 - 10 cos t N (B) 20 + 30 sin t N
(C) 20 + 30 cos t N (D) 50 - 30 cos t N
Solution 7: 53(a) k = = 500 N/m
54(b) time period T = 2 =2 =0.397 s
55(c)frequency f = = 2.51 Hz
angular frequency= 15.8 rad/s
= 2f
56(c) In equilibrium position acceleration = 0
k0 – mg = 0
0 = m
amplitude = maximum displacement from the equilibrium position
A = I 0 = 0.10 – 0.04 = 0.06 m.
57(d) Initial velocity = 0 (given)
Initial acceleration = = = 15 m/s2 upward.
58(a) Maximum velocity =A
= 0.06 15.8 = 0.95 m/s
Maximum acceleration = A2 = 15 m /s2
59(b) At one extreme position
spring force Fs = kI = 500 (0.1) = 50 N
At the other extreme position
Fs = k(A - 0) = 500 (0.06 – 0.04) = 10 N compression.
The equation of motion in SHM is
x = A sin (t + )
dx/dt = A cos (t + )
initial condition t = 0, x = 10 – 4 = 6 cm = 0.06 m.
A sin = 0.06
And A cos = 0
= and A = 0.06 m
x = 00.06 sin (t + /2)
60(c) At the given position x = A/2 = 0.03 m
0.03 = 0.06 sin (t + /2)
t = /3 t = /3 = 0.066 s
The instantaneous velocity
v = =
= A = 0.82 m/s upward
or v = A cos (t + /2)
= A cos (/3 + /2)
= A cos (5/3) = A = 0.82 m/s upward]
Instantaneous acceleration
a = - 2x= (A/2) 2 = 7.5 m/s2 i.e. 7.5 m/s2 upwards
spring force Fs = k (0 + x)
= mg + kA sin (t + )
= 20 + 30 sin (t + /2)
= 20 + 30 cos t N
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