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.

Elastic Potential Energy

When you elongate or contract a rod or wire by exerting a force, you do work. This work is stored as elastic potential energy in the rod or wire. The elastic potential energy per unit volume is easy to remember and is equal to ½ stress×strain. Since the Young’s modulus Y = stress/strain, we can modify the above expression as ½ Y×strain². Another form of the same expression is ½ (stress)²/Y.

Consider the following M.C.Q. which appeared in the Kerala Engineering Entrance test paper of 2006:

A work of 2×10^-2 J is done on a wire of length 50cm and area of cross section 0.5mm². If the Young’s modulus of the material of the wire is 2×10¹⁰ N/m², then the wire must be

(a) elongated to 50.1414cm (b) contracted by 2mm (c) stretched by 0.707mm (d) of length changed to 49.293cm (e) of length changed to 50.2cm

Since the total work done is involved in this prblem, we write,

½ (Y×strain²) ×volume = 2×10^-2

Here, strain = δ/0.5 where ‘δ’ is the elongation (or contraction) and volume = AL = (0.5×10^-6) ×0.5

Substituting proper values, δ works out to 1.414×10^-3 m = 1.414mm = 0.1414cm. The wire therefore elongates to 50.1414cm [option(a)].

An expression you should remember for obtaining the total work done in stretching or contracting a rod or wire through ‘l’ by exerting a force ‘F’ is, W = ½ F × l.

Breaking Strength and Breaking Stress

The breaking strength of a given wire depends on its area of cross section where as the breaking stress is a constant for a given material. Consider the following M.C.Q.:
A cable of length 10m and diameter 2cm can support a maximum load of 800 kg. If the length and the diameter of the cable are reduced to 5m and 1cm respectively, it will be able to support a maximum load of
(a) 200kg (b) 400kg (c) 800kg (d) 1600kg (e)100kg
The breaking strength is independent of the length but is directly proportional to the area of cross section. This follows from the expression for Young’s modulus (Y):
Y = Stress/strain. It is the strain which determines the breaking point of a cable.The strain at the breaking point is a constant for a given material. Since the Young’s modulus is a constant for a given material, the breaking stress also is a constant for a given material. Since the breaking stress is the ratio of the breaking strength to the area of cross section, it follows that the breaking strength is directly proportional to the area of cress section.
In the above problem, the breaking strength of the cable of diameter 2cm is 800 kgwt or(800×g) newton. Since the breaking stress is constant for the material of the cable, we can write, (800×g)/ (π×0.02^2) = mg/(π×0.01^2) where ‘m’ is the load the cable of half the diameter can support. Therefore, m = 200kg [option (a)]. It will be more convenient to write, 800/2^2 = m/1^2 to find ‘m’.
Suppose the breaking stress of a steel cable is ‘s’. What is the breaking stress of a steel cable of double the length and three times the diameter?
The answer, as you know, is ‘s’ since the breaking stress is a constant for a given substance.
Here is a very simple question which may mislead you if you are overconfident:
The Young’s modulus of a piano wire is Y. The Young’s modulus of another piano wire of half the thickness and twice the length is
(a) 2Y (b) 4Y (c) 8Y (d) Y/8 (e) Y
The correct option is (e) since the Young's modulus of a given substance is a constant.

You may visit physicsplus.blogspot.com for more multiple choice questions with solution.

Two Questions from Modern Physics

(1) The energy of a photon is 20eV. Its momentum in kg m/s is
(a) 2.56×10^-27 (b) 5.33×10^-27 (c) 1.066×10^-26 (d) 2.13×10^-26 (e) 3.18×10^-26
Since E=mc^2, momentum of the photon, p = E/c = (20×1.6×10^-19)/(3×10^8). Note that we have converted the energy in eV into joule. The answer is 1.066×10^-26 kg m/s given in option (c).
(2) The wave length associated with an electron having kinetic energy 6eV is
(a) 9A.U. (b) 5A.U. (c) 2.5A.U. (d) 1.5A.U. (e) 0.5A.U.
In the case of electrons, de Broglie wave length, λ = √ [ 150/V] A.U. where V is the accelerating voltage for the electron (= 6 volt since the energy is 6 eV).
Therefore, λ = √25 = 5 A.U.

Two Questions from Properties of Matter:
The following question appeared in the Kerala Engineering Entrance test paper of 2006:
The pressure inside two soap bubbles is 1.01 and 1.02 atmosphere respectively. The ratio of their respective volumes is
(a) 2 (b) 4 (c) 6 (d) 8 (e) 10
To the surprise of this author, a comparatively bright student omitted this question, which is quite simple. You know that the excess of pressure inside a bubble is 2T/r, where T is the surface tension and r is the radius. The ratio of the excess pressures inside the bubbles is P1/P2 = r2/r1. But, P1/P2 =0.01/0.02. [Since the actual pressures are 1.01 and 1.02 atmosphere]
So, we have r2/r1 = 1/2. The ratio of volumes, V1/V2 = (r1/r2)^3 = 2^3 = 8.
Consider now the following question which appeared in Kerala Medical Entrance test paper of 2006:
To what depth below the surface of sea should a rubber ball be taken so as to decrease its volume by 0.1%? [Take: Density of sea water = 1000kg/m^3, bulk modulus of rubber = 9×10^8 N/m^2, acceleration due to gravity = 10m/s^2]
(a) 9m (b) 18m (c) 180m (d) 90m (e) 900m
We have bulk modulus B = P/(dv/v) so that P = B(dv/v) = (9×10^8) ×0.001 = 9×10^5 pascal.
Therefore, hdg = 9×10^5 from which, h = (9×10^5)/(1000×10) = 90m. [Option (d)]