A small heavy spherical bob connected, to one end of a light inextensible and flexible string, with its other end fixed to a rigid and unyielding support, constitutes a simple pendulum.
Let P be the position of the bob at any instant, when the string makes an angle T, with the vertical.
(Figure. 15.14). If ‘l’ be the length of the string then, the net unbalanced force on the bob is F = mg sin T. [where m is the mass of the bob].
l
Assuming T to be very small sin T | T Net unbalanced force F = mg (T)
Now, if ‘x’ be the displacement of the bob from the mean position C, then T = x l . F m g x
l
§ ·
? ¨¨ ¸¸
© ¹
&
)&
Acceleration F g x
f
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fD x
? , which proves that the bob executes SHM for small amplitudes (less than 40).
Comparing with f Z2x,
We get 2 g g
l l
Z Z .
Alternatively: The torque acting about the point of suspension, is T = mg l sin T Now, moment of inertia I of the system about the point of suspension is I = ml2
From T = ID, we get 2 sin
Here ‘l’ is the “effective length” of the simple pendulum, measured from the point of suspension to the center of gravity of the bob.
A Few Important Facts Regarding Simple Pendulum
(a) Change of time period T due to a change in effective length.
[Law of lengths] T D l [other things remaining constant]
(i) If a child sitting on a swing stands up, then the effective length l decreases, due to an uplift of the center of gravity of the system. This reduces the time period consequently.
(ii) If a hollow sphere filled with a liquid, and having a hole at its base, be made the bob of a simple pendulum, the time period, initially increases, due to an increase in the effective length as the center of gravity gets lowered due to the outflow of liquid from the sphere. The time period once again, gets restored, when the whole liquid flows out, due to the fact that the center of gravity once again, comes to the center of the sphere.
(iii) If the string be elastic with Young’s modulus Y, then the length of pendulum will be l + ' l [where l is the original length and ' l, increase in length]
From Stress M g l
Now, since Mg < < A Y M g 1
? AY
2 l 1 M g
T S g §¨ AY ·¸
© ¹. [From Binomial expansion]
Obviously, the time period increases, in this case.
(iv) If the effective length be comparable to the radius of earth (Fig. 15.15)
The net unbalanced force on the bob = mg sin T I
, indicating that the bob, still executes SHM and its time period is not
infinity. Comparing the above equation with std. form 2 1 1 g l R
Putting l = R, we can determine, the corresponding time period as 2 1 2
? Neglecting 1
R in comparison to 1 l.
2 l
T S g
? as derived earlier..
(b) Change of time period with a change in acceleration due to gravity. (Law of gravity) T 1
g
D [other things remaining constant]
(i) If a simple pendulum be taken to higher altitudes, its time period increases, due to a decrease in the value of g. Consider a place P at height ‘h’ above the earth’s surface. (Fig 15.16). The value of g at that place g’ (say) is given by :
2
' g R 2
g
Rh
Where g = acc. due to gravity on the earth’s surface.
Time period 2
(ii) If a simple pendulum be taken to a place below the earth’s surface, the time period increases as a consequence of decrease in the value of acceleration due to gravity. Consider a place P, at a depth ‘d’
below the earth’s surface (Fig. 15.17) the acc. due to gravity at P is given by :
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' R d g g
R
§ ·
¨ ¸
© ¹ [where g is the acc. due to gravity on the earth’s surface]
Now, time period
2 '
T l S g
2 l R
g R d
S .
(iii) If a simple pendulum at equator, be shifted to poles, the time period decreases on account of an increase in the value of ‘g’. (acc. due to gravity is more at poles compared to equator).
(iv) The time period of a simple pendulum in a satellite or within a freely falling lift is infinite (or the pendulum does not oscillate due to the absence of any restoring force tending the bob to bring back to the mean position), on account of the fact that the value of g is zero.
(v) For a simple pendulum oscillating in a lift moving up with an acceleration a or down with a retardation a the effective value of g (say g’) will be g’ = g + a
2 l
T S g a
?
The time period always decreases.
For a simple pendulum, oscillating in a lift moving down with an acceleration a moving up with a retardation a, the effective value of g (say g’) will be g’ = ( g – a) down if g > a or (a – g) up if a > g.
(Fig. 15.18).
For a simple pendulum oscillating in a vehicle accelerating with an acceleration ‘a’, the time period decreases due to an increase in the value of ‘g’.
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Let us first determine the equilibrium position of the string (see figure 15.20). If Dbe the angle made by string with vertical. Then : T cos D = mg and T sin D = ma
D
D
Dividing tan a
D g D tan1§ ·¨ ¸ag
© ¹
(Opposite to the direction of acceleration of vehicle).
Effective external force = m a2g2
? Effective acceleration g' a2g2 1/ 2
? Time period 2 2 2 2 1/ 2
'
l l
T g g a
S S
.
(vi) If the bob of simple pendulum is acted upon by additional forces of electrical, magnetic, buoyant, gravitational origin, the time period will increase or decrease according as the net acceleration in the downward direction is decreased or increased (see problem No. 18, 19)
(C) Law of Mass: The time period of a simple pendulum is independent of the mass of the bob, other things remaining constant. For example, if two pendulums, be constructed of same effective length, but whose one of the bobs is solid and the other hollow; both oscillate with the same time period.
(d) Law of Isochronism: The time period of a simple pendulum is constant for every oscillation provided the amplitude remains small.
Example 17
Show that the maximum tension in the string of a simple pendulum, with angular amplitude T0is given by mg 1T02 . Hence find the maximum mass of bob that can be used, if the breaking load of the string be F.
Solution
Evidently, the tension is maximum, at the lowest position of the bob. (Fig. 15.21)
2 max max
T m g m v
l … (1)
Where vmax is the maximum velocity of the bob (attained in the mean position).
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vmax aZ
A simple pendulum of length l, is made to oscillate with a solid sphere (density D), in a non viscous fluid of densityU; find the period of oscillation (D > U )
Solution
Net force on the bob will be (D -U) Vg (down). Net Acceleration downward (See Fig. 15.22)
U
? The time period increases, due to a decrease in effective value of g.
Example 19
A simple pendulum consists of a small sphere of mass m, suspended by a thread of length L. The sphere carries a negative charge ‘q’. The pendulum is placed in a uniform electric field of strength E directed (i) vertically upwards (ii) vertically downwards, and (iii) horizontal. Find the time period in each case.
Solution
(a) The net force downward F = mg + qE Net acceleration downward F q E
f g
Time period
2 L
T q E
g m
S § ·
¨ ¸
© ¹ .
Note: The electric force experienced by a body carrying a negative charge is opposite to the direction of electric field.
(b) The net force downward = mg – qE [Assuming mg > qE] [Fig. 15.24 (a)]
? Acceleration qE
f g m
§ ·
¨ ¸
© ¹
Time period 2 L
T q E
g m
S § ·
¨© ¸¹
The net force upward = qE – mg [Assuming mg < qE] [Fig. 15.24 (b)]
Acceleration q E
f g
m
§ ·
¨ ¸
© ¹
?Time period
2 L
T q E
m g S §¨ · ¸
© ¹
(c) If ' 'T be the angle made by the string with the vertical (Fig. 15.25) Then T cos T = mg
T sin T = qE
tan q E tan 1 q E
mg m g
T T § ·
¨ ¸
© ¹
Net force on the bob = mg 2 qE 2 .
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? Net acceleration
A simple pendulum is suspended from the ceiling of a carriage, moving along a horizontal curve of radius of curvature r, with a speed v. If ‘l’ be effective length of the pendulum find (i) the tension in the string, while oscillating in the mean position and (ii) the period of oscillation (Assume T0as the angular amplitude).
Solution
Let us adopt the carriage as the frame of reference. Therefore, in addition to the tension and weight acting on the bob, a pseudo (centrifugal) force should also be applied on it. (Fig. 15.26)
At the mean position, the pendulum is stable.
? For its equilibrium net vertical and horizontal forces should be zero.
? T cos D = mg
Now, let T be the displacement (from mean position). at any instant Then, the net unbalanced force tangential to the motion of the bob.
F = mg sin D T - ma cos D T .
Where ‘x’ is the displacement of the bob from the mean position
ª § § &· § § &···º
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sin cos / cos sin
Now, maximum velocity (vel. at the mean position) vmax T0l Z
Or,
Now, from the FBD of the pendulum in its mean position, it is evident that, for equilibrium along the string.
Example 21
A simple pendulum, of length l, installed between two inclined walls, each making an angle D with the vertical, oscillates with a maximum speed of u, colliding elastically with the walls. Find the time period of oscillation. For the above problem, how is D related to u? (Fig. 15.28)
Solution
Had there been no walls, the linear amplitude of oscillation would be
[ max ]
a u From v aZ
Z
But since, g Z l a u l
g
Now, let the equation of motion for the pendulum be T T0 sinZt [where T0= angular amplitude]
The time taken by the pendulum to move from O (mean position) to A (extreme position), can be
obtained as u sin g l t D g l §¨¨© ·¸¸¹
sin g g l
l t u
D g sin 1 g l
t l u
§D ·
¨ ¸
¨ ¸
© ¹
sin 1 g l t l
g u
§D ·
¨¨© ¸¸¹
Evidently, the period of oscillation T = 4t 4 l sin 1 g l
T g u
§D ·
¨ ¸
¨ ¸
© ¹
Now, since, the bob collides with walls, the angular amplitude should exceed the value D . i.e.,
u or u g l
g l !D !D This is the required relation.
Note: If the bob just misses the collision with wall. Then, in the limiting case, angular amplitude = D
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Then , 4 l sin (1)1 This proves the validity of above derivation.
Second’s Pendulum: It is a pendulum with period of oscillation 2 sec or period of swing 1 sec. Length of a second’s pendulum at mean sea level, may be calculated as follows.
From
Note: A second’s pendulum always bears a time period of 2 sec. However, the length changes with a change in acceleration due to gravity.
Loss or Gain of Time by Pendulum Clocks:
Pendulum clocks gain time, if the time period of the pendulum decreases (due to a decrease in length in winter or due to increase in acc. due to gravity, as when taken from equator to poles). For example, let the time period of a second’s pendulum (originally) decrease to 1.99 sec. Now, each time. It oscillates, it counts 2 sec rather than 1.99 s. Thus, after a considerable no. of oscillations, the time counted or shown by the clock is more than the actual time, and consequently, is goes fast.
Similarly, Pendulum clocks loose time and go slow, when their time period increases.
Gain or loss of time due to a change in surrounding temperature:
Let T be the time period of clock pendulum of effective length l, showing the correct time.
Then, 2 l
T S g … (1)
Now, let ' be the change in temperature of the surrounding, the new length will beT
' 1
l l 'D T [where D is the coefficient of linear expansion of the material of the pendulum]
? The changed time period ' 2 l'
T S g 2 l 1
g
S 'D T … (2)
Dividing equation (2) by (1).
' 1/2
© ¹ [Expanding by binomial expansion]
1 1
i.e., Relative change in time period 1 2D 'T
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or, time lost or gained each second is 1 2D ' .T
Note: Unless otherwise stated, adopt the time period of clock pendulum as 2 sec.
Example 22
A clock with an iron pendulum keeps correct time at 200 C, how much time will it loose or gain, in a day, if the atmospheric temperature changes to 400 C. (Dfor iron = 0.000012/0 C)
Solution
The relative change in time (i.e., time lost or gained per swing i.e., in 1 sec is given by)
6 5
The pendulum of a certain clock has a time period 2.02 sec. How fast or slow does the clock run in an interval of one week.
Solution
Since the correct time period for a clock pendulum is 2 sec. Therefore, for every 2.02 sec. (one oscillation period) of actual time elapsed, it would count only 2 sec. In other words, as successive oscillations are repeated it goes on loosing time, and would run slow.
Loss of time per oscillation ( = 2 sec). = 2.02 – 2 = 0.02 sec.
? Loss of time per sec = 0.01 sec.
? Time lost in 1 week = 0.01 × 86400 × 7 = 6048 sec = 108 min.
Example 24
The linear displacement ‘x’ of a simple pendulum’s bob from its mean position varies as x = a sin (0.
707 St) where a is a constant and t is in sec. Find its length (take g = S2 m/s).
Solution
Since the simple pendulum executes SHM. Therefore comparing the displacement equation with standard form x = a sin Zt , we get
A simple pendulum of 1 meter, suspended from the ceiling of a lift executes oscillation with a period of 6.25 sec. If the lift is moving up with an acceleration ‘a’ find a (take g = S2m/s2).
Solution
If ‘a’ be the acceleration of the lift upwards then, net acceleration on the bob downwards, relative to the lift is g + a
Thus, the lift is retarding with a retardation of 9 2 2 25S m s/ Objective Questions:
1. The time period of a simple pendulum is independent of : Ans: (d).
a. The length of the string b. The radius of the bob c. Amplitude of oscillation d. None of these.
2. The time period of a simple pendulum of infinite length suspended on earth’s surface is :Ans: (c) a. Inifinte b. 59.8 minutes c. 84.6 minutes d. 24. hrs.
3. The ampliutude of oscillation of a simple pendulum decays exponentially with time due to air resistance and friction at the point of suspension. The total energy of the pendulum. Ans: (A)
a. Decays exponentially with time b. Decays linearly with time c. Decays at a steady rate
d. Decays at a rate less than the amplitude.
4. A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom of the sphere, the frequency of oscillation will : Ans: (D)
a. Go on increasing b. Go on decreasing
c. First increases and then decreases d. First decreases and then increases.
4. The formula 2 l
T S g holds good only for small angular amplitude T0 40 . For large amplitudes
2 l
The effective length l is the distance of the point of suspension O’ form the center of gravity (C.G) of the bob. As, water flows out, of the hole at the bottom, the C.G descends from center towards the bottom, increasing the effective length and consequently, f decreases.
However, when all the water has flown out, the C.G. of a hollow sphere is once again at its center and hence the effective length would decrease, thereby increasing the frequency.
5. A simple pendulum oscillates slightly above a large horizontal metal plate. The bob is given a charge.
The time period: Ans : B
a. Has no effect, whatever be the nature of charge.
b. Always decreases, whatever be the nature of charge.
c. Always increases, whatever be the nature of charge.
d. May increase or decrease depending upon the nature of charge
11. When any charge is given to the bob of the pendulum, it induces opposite charge on the metal plate, and hence a net force of attraction acts on the bob. Thus, the effective value of g increases.
6. The time period of a simple pendulum is T. In which of the following situations will the period increase?
Ans : B a. The pendulum is suspendend in lift, going with an acceleration (< g)
b. The bob is given a negative charge, while the point of suspension a positive charge.
c. The pendulum is suspended in a cabin, moving along a circle with uniform speed”
d. The pendulum is suspended in a cabin, moving along a straight path, with a uniform acceleration.
15. The electrostatic force of attraction acting on the bob is up, which decreases the net force of gravity downwards.
geff g
? .
So, since 1
eff
TD g a decrease in g increases the time period.
In the rest of all the cases, the effective value of g increases thereby decreasing the value of T.
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35. The bob of a simple pendulum of mass m executes oscillations with a total energy E. At any instant is at one of the extreme positions. Its linear momentum, after a phase change of
6
S rad takes place, will
be:
a. 2 m E b.
2
m E c. m E d. 3
4 m E .
36. A simple pendulum of length l has a time period T for small oscillations. A fixed obstacle is placed directly below the point of suspension, so that only the lower quarter of the string continues oscillations (see figure 15.74). The pendulum is released from rest at a certain point O. The period of oscillation, assuming small angles will be:
3 /4
O
a. 3 4
T b.
4
T c. 1 3
2
T d. T.
38. Two simple pendulums of time periods 3s and 5s respectively start oscillating simultaneously from their mean positions. What is the minimum time elapsed, before which they oscillate with the same phase?
a. 15 s b. 2 s c. 7.5 s d. 8 s.
39. In the above problem no. 38. if the bobs of the pendulums start their oscillations, from two opposite extreme positions, then, the answer would be :
a. 3.75 s b. 7.5 s c. 4 s d. 1.5 s.
40. Figure 15.75 shows three physical pendulums consisting of identical spheres rigidly connected by a mass less stiff rod. The pendulums are vertical and are made to oscillate about the point O. If their respective frequencies be f1, f2 and f3 then which of the following is correct :
a. f1 f2 f3 b. f1 f2! f3 c. f1 f2 f3 d. f1! f2! f3.
Answers: 35. a 36. a, 38. c 39. a 40. a.
Only one option is correct
17. A simple pendulum has time period T = 2 s in air. If the whole arrangement is placed in a nonviscous 1
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a. 2
2 s b. 4 s c. 2 2 d. 4 2 s.
21. An accurate pendulum clock is mounted on the ground floor of a high building. How much time will it lose or gain in one day if it is transferred to top storey of a building which is h = 200 m higher than the ground floor. Radius of earth is 6.4 × 106 m.
a. it will lose 6.2 s b. it will lose 2.7 s.
c. it will gain 5.2 s d. it will gain 1.6 s.
23. A simple pendulum 4 m long swings with an amplitude of 0.2 m. What is its acceleration at the ends of its path? (g = 10 m/s2)
a. zero b. 10 m/s2 c. 0.5 m/s2 d. 2.5 m/s2.
32. The maximum tension in the string of a pendulum is two times the minimum tension. Let T0 be the angular amplitude. Then cos T0 is :
33. A pendulum has time period T for small oscillations. An obstacle P is situated below the point of suspension O at a distance 3
4
l . 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
35. Time period of a simple pendulum of length L is T1 and time period of a uniform rod of the same length L pivoted about one end and oscillating in a vertical plane is T2. Amplitude of oscillations in both the cases is small. Then T1/T2 is:
a. 4
3 b. 1 c. 3
2 d. 1
3 .
63. The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle which moves without friction, down an inclined plane of inclination D, is given by :
a. 2
66. A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y = Kt2, (K = 1 m/s2) where y is the vertical displacement. The time period now becomes
T2
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a. 6
5 b. 5
6 c. 1 d. 4
5. 15.9 Physical Pendulum or Compound Pendulum:
Any rigid body, suspended about a fixed point and capable of oscillating about a horizontal axis passing through the point, constitutes a physical or a compound pendulum. Figure 15.45 along side shows a rigid body (of any shape) suspended about a fixed point O (the center of suspension) capable of rotating about a horizontal axis through O. Let G be the center of gravity at a distance l from the point of suspension O. If ‘R’ be the vertical reaction upwards at the support O, then for vertical equilibrium, we have.R = mg
l
When at rest, the line of action of R and mg is the same and hence, there is no net torque on the body.
If the body be given a slight angular displacement about O, it starts oscillating, the oscillations being angular SHM.
Consider the position of the body; at any instant (during oscillatory motion) when the line joining O and G makes an angle ' 'T with the vertical .
The net torque on the body about the point O will be given by W m g lsinT If I be the moment of inertia of the body about O, then
2
T = angular acceleration]
2
[The – ve sign, here indicates that the direction (or sense] of T and W are opposite]
[ for small angles sinT = T]
Which reveals the fact that, the body performs angular SHM with angular frequency m g l
Z I
Now, if T be the time period for each oscillation, then
2 2 I
T m g l
S S
Z ... (20).
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If K be the radius of gyration of the body about a parallel axis through the center of gravity then, I = mK2 + ml2
2 2
2 K l
T S g l . … (21)
If L be the length of an equivalent simple pendulum (i.e., a pendulum whose time period is same as that of the compound pendulum)
Then , 2 L
T S g
Comparing the above equation with equation (21) we get K2
L l
l .
Thus, the length of equivalent simple pendulum is K2
l more than the distance between the point of suspension O and the center of gravity G. If we choose a point O’, on the line OG produced such that GO’ =
K2
l , then O’ is known as the center of oscillation.
Interchangeability of Center of Suspension and Center of Oscillation:
Suppose we shift the point of suspension from O to O’. Then time period
2 2
Where l’ is the distance of O’ (new point of suspension) from center of gravity G.
But,
2 2
Thus, it is evident that, the time period remains the same for both the center of suspension and center of oscillation i.e., both are interchangeable.
Minimum Period of Oscillation:
From equation (21), the time period T for a compound pendulum is given by
2
Which can be rewritten as
2 2 square can be zero.
? K = l
Thus, the time period of oscillations is minimum when the distance of the center of suspension from the center of gravity is same as the radius of gyration of the body about a parallel axis through C.G.
Also, min 2
2 k
T S g ... (22)
Example 33
A uniform circular disc of radius R, is suspended about a point (not its center) such that it can oscillate about a horizontal axis through that point. What is the distance of this point from the center, for the
A uniform circular disc of radius R, is suspended about a point (not its center) such that it can oscillate about a horizontal axis through that point. What is the distance of this point from the center, for the