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Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
What are the Uses of Simple Pendulum?
Definition: A simple pendulum of a time period of 2 seconds, or a half time period of 1 second, is called a seconds pendulum.
A seconds pendulum has a time period. T = 2s.
It is known, T = 2π\(\sqrt{\frac{L}{g}}\) or, 2 = 2π\(\sqrt{\frac{L}{g}}\) or, l = π\(\sqrt{\frac{L}{g}}\)
or, L = \(\frac{g}{\pi^2}\) = \(\frac{981}{\pi^2}\) = 99.40 cm = 0.9940 m.
Pendulum clock running fast or slow: A pendulum clock marks time by means of its time period. From the equation T = 2π\(\sqrt{\frac{L}{g}}\), it can be shown that if
1. the effective length L of a pendulum changes, or
2. the value of the acceleration due to gravity changes, the time period of a seconds pendulum does not remain 2 s, but increases or decreases accordingly. Increase in the time period means that the pendulum oscillates slowly, i.e., the clock goes slow. On the other hand, a decrease in the time period makes the clock run fast.
For example, when a pendulum clock is taken from the poles to the equator, or from the sea level to the top of a mountain, or from the earth’s surface to deep inside a mine, the acceleration due to gravity g decreases; hence, the time period increases. Thus, in each case, the clock goes slow. A pendulum clock may run slow or fast depending on the temperature. At a higher temperature, due to the expansion of the metallic suspender and the bob, the effective length increases and the time period also increases. On the other hand, at low temperatures. the time period decreases. Therefore, a pendulum clock runs slow in summer and fast in winter. To get the correct time from the same pendulum at different temperatures, compensated pendulums are used.
Some Uses of a Simple Pendulum
Finding the value of g: From equation (1) in Section 1.4.2 we get,
g = 4π2\(\frac{L}{T^2}\) ……… (1)
Using this equation, value of acceleration due to gravity (g) of a place can be determined. For different lengths of strings, the time period (T) and the effective length (L) of a pendulum can be evaluated by experiment. Then \(\frac{L}{T^2}\) value in each case is determined and average value of L is calculated. Substituting this average value in equation (1), the value of g can be obtained.
Determining the height of a place: Let R = radius of earth, h = height of a place (say, top of a hill), g = value of acceleration due to gravity on earth surface, g’ = value’ of acceleration due to gravity at height h, T = time period of a simple pendulum at a fixed place on earth surface and T’ = time period of same pendulum at height h.
From equation (1) in Section 1.5.2 in the chapter Newtonian Gravitation and Planetary Motion,
Finding the depth of a mine: Let h = depth of mine, g = value of acceleration due to gravity on earth surface, g’ = value of acceleration due to gravity at the depth h below earth surface, T = time period of a simple pendulum on earth surface, T’ = time period of same pendulum at depth h.
From equation (4) in Section 1.5.3 in the chapter Newtonian Gravitation and Planetary Motion, we get
Knowing the value of R and determining the values of T and T’ using a stopwatch, the value of h can be determined from equation (7).
Numerical Examples
Example 1.
A simple pendulum executes 40 complete oscillations in a minute. What is the effective length of the pendulum? g = 980 cm ᐧ s-2.
Solution:
Time period (T) = \(\frac{60}{40}\) = \(\frac{3}{2}\)s
Now, T = 2π\(\sqrt{\frac{L}{g}}\) or T2 = 4π2\(\frac{L}{g}\)
or, L = \(\frac{g T^2}{4 \pi^2}\) = \(\frac{980 \times\left(\frac{3}{2}\right)^2}{4 \times \pi^2}\) = 55.9 cm
Example 2.
What will be the percentage increase in the time period of a simple pendulum when its length is increased by 21%?
Solution:
Given, increase in length of the pendulum = 0.21 L, where L is the initial length.
Hence, increased length L’ = L + 0.21 L = 1.21 L
Let, the time period change to T’ from T due to the change in length.
Increase in time period = T’ – T = 1.1 T – T
= 0.1T = 10% of T
Hence, the time period increases by 10%.
Example 3.
Two simple pendulums of lengths loo cm and 101 cm are set into oscillation at the same time. After what time does one pendulum gain one complete oscillation over the other?
Solution:
Length of the first pendulum is comparatively less, and hence, its time period is also less; thus the first pendulum oscillates faster. By the time the second pendulum executes n oscillations, suppose the first one completes (n + 1) oscillations. Hence, if T1 and T2 are time periods of the first and the second pendulums,
(n + 1)T1 = nT2 or, \(\frac{T_2}{T_1}\) = \(\frac{n+1}{n}\) = 1 + \(\frac{1}{n}\) ……… (1)
Thus, the required time = time of 201 complete oscillations of the first pendulum = 201 × T1
= 201 × 2π\(\sqrt{\frac{L_1}{g}}\)
= 201 × 2 × π\(\sqrt{\frac{100}{980}}\) ≈ 403s = 6 min 43 s.
Example 4.
Find the length of a simple pendulum on the surface of the moon that has a time period same as that of a simple pendulum on the earth’s surface. Mass of earth is 80 times that of moon and the radius of earth is 4 times that of moon.
Solution:
If masses of the earth and the moon are M1 and M2 respectively, \(\frac{M_1}{M_2}\) = 80
Again, their radii are R1 and R2 (say).
So, \(\frac{R_1}{R_2}\) = 4
Acceleration due to gravity on earth, g1 = \(\frac{G M_1}{R_1^2}\), acceleration due to gravity on moon, g2 = \(\frac{G M_2}{R_2^2}\)
Also, in the case of simple pendulum,
time period on earth’s surface T = 2π\(\sqrt{\frac{L_1}{g_1}}\)
and time period on the moon’s surface T = 2π\(\sqrt{\frac{L_2}{g_2}}\)
∴ \(\sqrt{\frac{L_1}{g_1}}\) = \(\sqrt{\frac{L_2}{g_2}}\) or, \(\frac{L_1}{L_2}\) = \(\frac{g_1}{g_2}\) = 5
∴ L2 = \(\frac{L_1}{5}\)
Hence, length of pendulum on the moon’s surface should be \(\frac{1}{5}\)th of its length on the earth’s surface.
Example 5.
Two pendulums of time periods 1.8 s and 2 s are set into oscillation at the same time. After how many seconds will the faster moving pendulum execute one complete oscillation more than the other? How many complete oscillations will the faster moving pendulum execute during this time?
Solution:
Suppose the faster pendulum executes one more oscillation than the other after t s.
Number of complete oscillations of the first pendulum in t s = \(\frac{t}{1.8}\)
Number of oscillations of the second pendulum in t s = \(\frac{t}{2}\).
According to the given condition, \(\frac{t}{1.8}\) – \(\frac{t}{2}\) = 1
or, 0.2 t = 2 × 1.8 or, r = 18 s
Number of complete oscillations executed by the faster pendulum (time period 1.8 s) = \(\frac{18}{1.8}\) = 10.
Example 6.
A pendulum clock runs 20s slow per day. What should be the change in length of the clock so that it records correct time? Take the pendulum as a simple pendulum. [WBJEE ’03]
Solution:
Half time period of a simple pendulum
t = \(\frac{T}{2}\) = π\(\sqrt{\frac{L}{g}}\)
For a perfect seconds pendulum t = 1 s
∴ 1 = π\(\sqrt{\frac{L}{g}}\) or, L = \(\frac{g}{\pi^2}\) …….. (1)
Also, 1 d = 24 × 60 × 60 = 86400 s. Number of half oscillations executed in a day by a pendulum which runs 20 s slow per day = 86400 – 20 = 86380.
Hence, half time period of that pendulum t1 = \(\frac{86400}{86380}\)s.
If the length of this simple pendulum is L1,
then t1 = π\(\sqrt{\frac{L_1}{g}}\) or, L1 = \(\frac{g t_1^2}{\pi^2}\) ……….. (2)
Subtracting equation (1) from equation (2),
L1 – L = \(\frac{g}{\pi^2}\) (\(t_1^2\) – 1) = \(\frac{980}{\pi^2}\)[\(\left(\frac{86400}{86380}\right)^2\) – 1]
= 0.046 cm = 0.46 mm
Hence, to get correct time, length of the pendulum is to be decreased by 0.46 mm.
Alternative Method:
Half time period of a perfect seconds pendulum = 1 s.
Number of half oscillations of a clock that runs slow by t0 s in a day = 86400 – t0.
Hence, half time period = \(\frac{86400}{86400-t_0}\)s
∴ Increase in the value of half time period,
dt = \(\frac{86400}{86400-t_0}\) – 1 = \(\frac{t_0}{86400-t_0}\)s
If the value of t0 is negligibly smaller than 84600,
dt = \(\frac{t_0}{86400}\)s
Now, t = π\(\sqrt{\frac{L}{g}}\) or, log t = logπ + \(\frac{1}{2}\)logL – \(\frac{1}{2}\)logg
Differentiating, \(\frac{d t}{t}\) = \(\frac{1}{2} \frac{d L}{L}\) – \(\frac{1}{2} \frac{d g}{g}\) ………… (1)
Putting values of t and dt in (1),
\(\frac{t_0}{86400}\) = \(\frac{1}{2}\left(\frac{d L}{L}-\frac{d g}{g}\right)\)
or, t0 = 43200\(\left(\frac{d L}{L}-\frac{d g}{g}\right)\) ……. (2)
This equation can be used as a rule for a seconds pendulum. For a seconds pendulum L = \(\frac{g}{\pi^2}\) and on the surface of the earth g = 980 cm ᐧ s-2. If the clock runs fast, value of is negative.
In the given problem, t0 = 20 s;
If there is no change in the value of g, dg = 0
∴ t0 = 43200 × \(\frac{d L}{L}\)
or, dL = \(\frac{20 \times L}{43200}\) = \(\frac{20}{43200}\) × \(\frac{980}{\pi^2}\) = 0.046 cm = 0.46 mm
Hence, length of the defective clock has increased by 0.46 mm. Thus to get correct time, its length needs to be decreased by 0.46 mm.
Example 7.
A pendulum of length 60 cm is suspended inside an aeroplane. The aeroplane is flying up with an acceleration of 4 m ᐧ s-2 making an angle of 30° with the horizontal. Find the time period of oscillation of the pendulum.
Solution:
Horizontal component of acceleration a of the plane = a cos30° and vertical component = a sin 30° [Fig.]
Hence, the downward acceleration experienced by the pendulum bob = g – (-a sin 30°) = g + a sin 30°
So, the acceleration of the pendulum bob
Example 8.
The effective length of a simple pendulum is 1 m and the mass of its bob is 5g. If the amplitude of motion of the pendulum is 4 cm, what is the maximum tension on the string to which the bob is attached?
Solution:
The velocity of the bob is maximum at its position of equilibrium. The maximum velocity vmax = ωA = \(\frac{2 \pi A}{T}\).
At this position, centripetal force is also maximum, whose value is Fc = \(\frac{m v_{\max }^2}{L}\) = \(\frac{m \omega^2 A^2}{L}\)
[L = effective length of the pendulum]
The resultant of the weight mg of the bob and tension F on the string becomes equal to this force F.
Example 9.
Prove that the change in the time period t of a simple pendulum due to a change ∆T of temperature is, ∆t = \(\frac{1}{2}\)αt∆T, where α = coefficient of linear expansion.
Solution:
If L is the effective length of a simple pendulum, then its time period is, t = 2π\(\sqrt{\frac{L}{g}}\)
For a change T of temperature, the length becomes, L’ = L(1 + α∆T)
Therefore the time period,
[neglecting the terms containing α2, α3, etc. since α is very small]
The change in time period, ∆T = t’ – t = \(\frac{1}{2}\)αt∆T (Proved)
Alternative Method:
t = 2π\(\sqrt{\frac{L}{g}}\)
logt = log 2π + \(\frac{1}{2}\)logL – \(\frac{1}{2}\)logg
Differentiating with respect to L,
\(\frac{1}{2}\)dt = \(\frac{1}{2L}\)dL
∴ \(\frac{d t}{t}\) = \(\frac{1}{2} \frac{d L}{L}\) …… (1)
Now Lt = L(1 + α∆T)
or, Lt – L = Lα∆T or, dL = Lα∆T
or, \(\frac{d L}{L}\) = α∆T ………. (2)
From equation (1) and (2) we get,
\(\frac{d t}{t}\) = \(\frac{1}{2}\)α∆T or dt = \(\frac{1}{2}\)α∆T ᐧ t (Proved).
Example 10.
The bob of a simple pendulum is made of brass and its time period is T. It is completely immersed in a liquid and is allowed to oscillate. If the density of the liquid is \(\frac{1}{8}\)th of the density of brass, what will be the time period of oscillation of the pendulum now?
Solution:
Initial time period of the simple pendulum,
T = 2π\(\sqrt{\frac{L}{g}}\) [L = effective length of the pendulum]
If m is the mass, V is the volume of the bob and d is the density of brass, then the apparent weight of the bob inside the liquid,
W1 = W – buoyancy (weight of the displaced liquid)
= Vdg – V\(\frac{d}{8}\)g = \(\frac{7}{8}\)Vdg
If g1 is the effective acceleration due to gravity in the immersed condition, then
W1 = mg1 = \(\frac{7}{8}\)Vdg or, g1 = \(\frac{7}{8} \frac{m g}{m}\) = \(\frac{7}{8}\)g
∴ Final time period, T1 = 2π\(\sqrt{\frac{L}{g_1}}\) = 2π\(\sqrt{\frac{8 L}{7 g}}\) = \(\sqrt{\frac{8}{7}}\)T.