The period of oscillation of a simple pendulum is given by the simple equation,
T = 2π√(L/g).
Questions based on this equation can be seen in test papers. The following question appeared in the I.I.T. screening test of 2005:
A simple pendulum of length ‘L’ has a small spherical bob of mass ‘m’ that carries a positive charge ‘q’. The pendulum is located in a uniform electric field ‘E’ directed vertically upwards. If the electric force is less than the gravitational force, the period of oscillation of this pendulum is
(a) 2π√(L/g) (b) 2π√[L/(g-E)] (c) 2π√[L/(g+Eq/m)] (d) 2π√[L/(g-Eq/m)]
(e) 2π√[L/(g+E)]
Here also you have to replace ‘g’ (in the expression for the period) by the net acceleration, as in the previous question. But the net acceleration in the present case is (g-a) where a = Eq/m, which is the acceleration produced by the electric force Eq. Therefore ‘g’ is to be replaced by (g-Eq/m). The correct option therefore is (d).
Note that the real weight of the bob is mg. The apparent weight of the bob is (mg-Eq) since the electric force is upwards. The net downward acceleration therefore is (g-Eq/m).
A simple pendulum will not work on an artificial satellite orbiting round the earth since the pendulum bob becomes weightless and hence there is no restoring force mgsinθ. But you can use a spring loaded with a mass as an oscillator even on an artificial satellite or for that matter, even in a region of space where there is no gravitational force. The period of oscillation of such a spring-mass system, as you might be remembering is
T = 2π√(m/k), where m is the mass and k is the spring constant.
This equation is devoid of g and so the system works even in weightless situations.
Now suppose that the bob of a simple pendulum of length ‘L’is immersed in a non-viscous liquid of density equal to one-tenth the density of the material of the bob. The apparent weight of the bob is now reduced to nine-tenth of the real weight(because of the upthrust of the liquid). The period of oscillation of the pendulum therefore increases as
T = 2π√(10L/9g)
[The above equation is easily obtained if you remember that the mass of the bob is vρ and the apparent weight of the bob is v(ρ-σ)g so that the net value of the downward acceleration of the bob is v(ρ-σ)g /vρ = (1- σ/ρ)g = (1- 1/10)g = (9/10)g.]
In the case of a spring-mass system, there is no change in the period if the oscillating mass is immersed in a non-viscous liquid, since the period is independent of ‘g’.
Simple Pendulum of Infinite Length
The period of oscillation of a simple pendulum, as you know, is given by
T = 2π√(L/g).
Questions based on this equation can be seen in test papers. The following question appeared in the I.I.T. screening test of 2005:
The point of suspension of a simple pendulum with normal time period T1 is moving upward according to the equation, y=kt2 where k=1 m/s2. If the new time period is T, the ratio T12/ T2 will be
(a) 2/3 (b) 5/6 (c) 6/5 (d) 3/2
This is a simple question. But you should note that the acceleration due to gravity ‘g’ is to be replaced by the net acceleration (g+a) since the pendulum as a whole is moving up with an acceleration ‘a’ which is obtained by differentiating the equation y = kt2 twice. Therefore, a = 2k = 2 since k=1. The new period is given by,
T = 2π√[L/(g+a)] = 2π√[L/(10+2)] = 2π√(L/12).
The normal period of the pendulum is
T1 = 2π√(L/10).
Therefore, T12/ T2 =12/10 = 6/5 [Option (c)]
Let us consider another question in which en electric force modifies the effective weight of the bob of the pendulum, thereby changing the period of oscillation:A simple pendulum of length ‘L’ has a small spherical bob of mass ‘m’ that carries a positive charge ‘q’. The pendulum is located in a uniform electric field ‘E’ directed vertically upwards. If the electric force is less than the gravitational force, the period of oscillation of this pendulum is
(a) 2π√(L/g) (b) 2π√[L/(g-E)] (c) 2π√[L/(g+Eq/m)] (d) 2π√[L/(g-Eq/m)]
(e) 2π√[L/(g+E)]
Here also you have to replace ‘g’ (in the expression for the period) by the net acceleration, as in the previous question. But the net acceleration in the present case is (g-a) where a = Eq/m, which is the acceleration produced by the electric force Eq. Therefore ‘g’ is to be replaced by (g-Eq/m). The correct option therefore is (d).
Note that the real weight of the bob is mg. The apparent weight of the bob is (mg-Eq) since the electric force is upwards. The net downward acceleration therefore is (g-Eq/m).
A simple pendulum will not work on an artificial satellite orbiting round the earth since the pendulum bob becomes weightless and hence there is no restoring force mgsinθ. But you can use a spring loaded with a mass as an oscillator even on an artificial satellite or for that matter, even in a region of space where there is no gravitational force. The period of oscillation of such a spring-mass system, as you might be remembering is
T = 2π√(m/k), where m is the mass and k is the spring constant.
This equation is devoid of g and so the system works even in weightless situations.
Now suppose that the bob of a simple pendulum of length ‘L’is immersed in a non-viscous liquid of density equal to one-tenth the density of the material of the bob. The apparent weight of the bob is now reduced to nine-tenth of the real weight(because of the upthrust of the liquid). The period of oscillation of the pendulum therefore increases as
T = 2π√(10L/9g)
[The above equation is easily obtained if you remember that the mass of the bob is vρ and the apparent weight of the bob is v(ρ-σ)g so that the net value of the downward acceleration of the bob is v(ρ-σ)g /vρ = (1- σ/ρ)g = (1- 1/10)g = (9/10)g.]
In the case of a spring-mass system, there is no change in the period if the oscillating mass is immersed in a non-viscous liquid, since the period is independent of ‘g’.
Simple Pendulum of Infinite Length
The period of oscillation of a simple pendulum, as you know, is given by
T=2π√(l/g), with usual notations. If the length of the pendulum is not negligible compared to the radius(R) of the earth, the period is given by
T = 2π√[Rl/(R+l)g]. This equation shows that in the case of a simple pendulum of infinite length(or, to be more realistic, in the case where the length is large compared to the radius of the earth), the period is
T = 2π√(R/g)
On substituting for R = 6400 km (=64×10^5m) and g = 9.8m/s^2, the period works out to be 5078seconds or, 84.6 minutes.
Remember the above equation. The period of oscillation of a stone dropped into an imaginary hole drilled along the diameter of the earth and the orbital period of a satellite moving close to the earth’s surface also are given by this equation.
T = 2π√[Rl/(R+l)g]. This equation shows that in the case of a simple pendulum of infinite length(or, to be more realistic, in the case where the length is large compared to the radius of the earth), the period is
T = 2π√(R/g)
On substituting for R = 6400 km (=64×10^5m) and g = 9.8m/s^2, the period works out to be 5078seconds or, 84.6 minutes.
Remember the above equation. The period of oscillation of a stone dropped into an imaginary hole drilled along the diameter of the earth and the orbital period of a satellite moving close to the earth’s surface also are given by this equation.
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