If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Showing posts with label simple pendulum. Show all posts
Showing posts with label simple pendulum. Show all posts

Friday, June 08, 2007

Kerala Medical Entrance 2007 Questions on Simple harmonic Motion

The following two questions were included from simple harmonic motion, in KEAM (Medical) 2007 question paper:

(1) A simple pendulum has a time period T in vacuum. Its time period when it is completely immersed in a liquid of density one-eighth of the density of material of the bob is

(a) √(7/8) T (b) √(5/8) T (c) √(3/8) T (d) √(8/7) T (e) √(8/5) T

The case of a pendulum immersed in a liquid was briefly discussed in the post dated 16th August 2006.

The mass of the bob is vρ where ‘v’ is the volume of the bob and ‘ρ’ is the density of the material of the bob. The apparent weight of the bob is v(ρ-σ)g where ‘σ’ is the density of the liquid so that the net value of the downward acceleration of the bob is v(ρ-σ)g /vρ = (1- σ/ρ)g = (1- 1/8)g = (7/8)g.

The period of oscillation of the pendulum suspended in vacuum (or, as usual, in air), is given by

T = 2π√(L/g) where 'L’ is the length and ‘g’ is the acceleration due to gravity.

When the pendulum bob is immersed in the liquid, the net acceleration is (7/8)g and the period becomes

Tliquid = 2π√(8L/7g) = √(8/7) T.

You may see the earlier post on simple pendulum containing other interesting questions, by clicking on the label ‘simple pendulum’ below this post or by making use of the search facility at the top of this page.

(2) A body of mass 20 g connected to a spring of constant K executes SHM with a frequency (5/π) Hz. The value of spring constant is

(a) 4 Nm–1 (b) 3 Nm–1 (c) 2 Nm–1 (d) 5 Nm–1 (e) 2.5 Nm–1

This is a standard direct question. The frequency of oscillation of a spring-mass system is given by

N = (1/2π)√(K/m) where K is the spring constant and ‘m’ is the mass.

Therefore, 5/π = (1/2π)√(K/0.02), from which

K = 2 Nm–1

Tuesday, August 29, 2006

Questions on Gravitation
You should definitely remember the following relations to ensure good score in gravitation:
(1) Acceleration due to gravity at a height ‘h’ is given by
g’ = GM/(R+h)2, with usual notations.
Surface value of acceleration due to gravity, g = GM/R2
If ‘h’ is small compared to the radius ‘R’ of the earth, g' = g(1-2h/R)
(2) Acceleration due to gravity at a depth ‘d’ is given by g'' = g (1-d/R)
Note that this is true for all values of ‘d’.
(3) Gravitational potential energy of a mass ‘m’ at a height ‘h’ is given by U= -GMm/(R+h)
This can be written as U = -GMm/r where ‘r’ is the distance from the centre of the earth.
(4) Escape velocity from the surface of earth (or any planet or star),
ve = √(2GM/R) = √(2gR)
Escape velocity from a height ‘h’ = √[2GM/(R+h)] = √[2g'(R+h]
(5) Kinetic energy and total energy of a satellite are equal in magnitude. But K.E. is positive where as total energy is negative. The potential energy of a satellite is negative and is equal to twice the total energy.( Note that this is true in all central field motion under inverse square law force, as for example, the energy of the electron in the hydrogen atom.)
In the case of a satellite of mass ‘m’ in an orbit of radius ‘r’:
Potential energy = -GMm/r
Kinetic energy = +GMm/2r
Total energy = -GMm/2r
(6) As per Kepler’s law, T2 α r3
(7) Orbital speed ‘v’ of a satellite in an orbit of radius ‘r’ is independent of its mass and is given by v = √(GM/r) = √(g'r) where g' is the acceleration due to gravity at the orbit and M is the mass of the earth (or planet).
Let us now discuss the following question which appeared in the Kerala Medical Entrance Test paper of 2002:
The escape velocity of a body on an imaginary planet which has thrice the radius of the earth and twice the mass of the earth is (where ve is the escape velocity on the earth)
(a) √(2/3).ve (b) √(3/2).ve (c) √(2).ve/3 (d) 2ve/√3 (e) 2ve/3
We have ve = √(2GM/R). Replacing R with 3R and M with 2M, we obtain the answer as√(2/3).ve [option (a)].
Consider now the following question which may confuse some of you:
The orbital velocity of an artificial satellite near the surface of the moon is increased by 41.4%. The satellite will
(a) move in an orbit of radius greater by 41.4% (b) move in an orbit of radius twice the original value (c) move in an elliptical orbit (d) fall down (e) escape into outer space
The correct option is (e). The orbital speed of any satellite moving round any heavenly body is √(GM/r) where as the escape velocity is √(2GM/r). This means that the escape velocity is √2 times the orbital speed or 1.414 times the orbital speed. Therefore, when the orbital speed is increased by 41.4% the satellite will escape into outer space.
Consider now the question which appeared in the Kerala Engineering Entrance Test paper of 2001:
The orbital speed of an artificial satellite very close to the surface of the earth is V0. Then the orbital speed of another artificial satellite at a height equal to 3 times the radius of the earth is
(a) 4V0 (b) 2V0 (c) V0 (d) 0.5V0 (e) 2V0/3
We have V0 =√(GM/R). At a height equal to three times the radius of the earth, the orbital velocity is obtained by replacing R with R+3R = 4R. The answer is 0.5R [option (d)].
The following simple question appeared in the IIT 2001 test paper:
A simple pendulum has a time period T1 when on earth’s surface, and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T2/T1 is
(a) 1 (b) √2 (c) 4 (d) 2
The required ratio is [2π√(L/g’)] / [2π√(L/g)] = √(g/g’). But g = GM/R^2 and g’ = GM/(2R)^2 so that g/g' = 4. The answer therefore is 2 [option(d)].
Consider now the following question which appeared in H.P.P.M.T.2005:
If a body of mass ‘m’ is raised from the surface of the earth to a height ‘h’ which is comparable to the radius of the earth R, the work done is
(a) mgh (b) mgh[1-(h/R)] (c) mgh[1+(h/R)] (d) mgh/[1+(h/R)]
Note that ‘g’ is the acceleration due to gravity on the surface of the earth. The work done for raising the body is the difference between the gravitational potential energies at the height ‘h’ and at the surface. Therefore, work done, W = -GMm/(R+h) – (-GMm/R) where M is the mass of the earth. Therefore, W = GMm/R - GMm/(R+h) = mgR – mgR/[1+(h/R)], on substituting g=GM/R2.
Thus, W = mgR[1- 1/1+(h/R)] = mgh/[1+(h/R)]

Wednesday, August 16, 2006

Questions on Simple Pendulum

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:

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.