Spring Constant and Potential Energy of Spring

Spring constant (force constant of a spring) is the force required for unit extension (or contraction) in a spring. Suppose a spring is cut into two pieces of equal length. Will the spring constant change? Don’t be doubtful. The spring constant will be doubled. If you cut a spring of spring constant ‘k’ into ‘n’ pieces of equal length, the spring constant of each piece will be nk.
Springs may appear in series and parallel combinations in certain questions, as for example, in problems involving the period of oscillation of a spring-mass system. If you have springs of constants k₁, k₂, k₃….etc. in series, the net spring constant ‘k’ of the combination is given by the reciprocal relation, 1/k = 1/k₁ + 1/k₂ + 1/k₃ + . . . . . . .etc.
If you have springs in parallel, the net spring constant ‘k’of the combination is given by k = k₁+ k₂+ k₃+ . . . . .etc.
Often you will encounter questions involving the potential energy of a spring. You should remember that the potential energy of a spring of constant ‘k’, stretched (or contracted) through a distance ‘x’ is (½) kx². Consider the following MCQ:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, its potential energy will be
(a) 3U (b) 6U (c) 9U (d) U (e) 36U
Since the potential energy is ½ kx², it follows that the P.E. is directly proportional to the square of the stretch (extension). The extension being 3 times, the P.E. must be 9 times. So, the correct option is (c).
Suppose the above question is asked in the following modified form:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, the increase in its potential energy will be
(a) 4U (b) 6U (c) 9U (d) 8U (e) 36U
Since the difference between the initial and final potential energies is required in this problem, the answer is 9U-U = 8U.
Now, consider the following question:
The tension in a spring of spring constant k is T. The potential energy of the spring is
(a) T²/k² (b) T²/k (c) 2T²/k (d) T²/2k (e) 2T²/k²
Potential energy of a spring, as you know, is (½)kx². But k=T/x from which x=T/k. On substituting this value of x in the expression for potential energy, we obtain option (d) as the answer.
Now, consider the following M.C.Q. which appeared in IIT screening 2002question paper:
An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially un-stretched. Then the maximum extension in the spring is
(a) 4Mg/k (b) 2Mg/k (c)Mg/k (d) Mg/2k
When the mass is released, its gravitational potential energy is decreased and the elastic potential energy in the spring is increased by an equal amount. If ‘x’ is the maximum extension produced (when the mass reaches the lowest position), we have Mgx = (½) kx², from which x = 2Mg/k. [Option (b)]

All India Engineering/Architecture Entrance Examination (AIEEE) 2007

Application Form and the Information Bulletin of the All India Engineering/Architecture Entrance Examination 2007 to be conducted by Central Board of Secondary Education (CBSE) on 29th April 2007 will be distributed from 1-12-2006 to 5-1-2007. The Information Bulletin and the Application Form can be obtained personally from designated branches of Syndicate Bank/Institutions and Regional Offices of CBSE. Designated branches of Syndicate Bank in Kerala are:

(1) Thiruvananthspuram- Syndicate Bank, Co-operative Agriculture and Rural Development Bank Building, P .B.No.314, Thiruvananthapuram-695001

(2) Ernakulam- Syndicate Bank, Pioneer Towers, First Floor, P.B.No.2616, Shanmugham Road, Ernakulam-682031

(3) Kozhikode- Syndicate Bank, P.B.No.61, Cherootty road, Kozhikode-673001

(4) Kottayam- Syndicate Bank,, P.B.No.121, Baker Junction, Kottayam-686001

(5) Trichur- Syndicate Bank, Palace Road, Trichur-680020

The Information Bulletin and the Application form can be obtained by post from the Joint Secretary (AIEEE), Central Board of secondary Education, 17, Rouse Avenue, Institutioal Area (Near Bal Bhavan), New Delhi-110002.

The cost of Information Bulletin containing the Application Form, inclusive of the examination fee for BE/B.Tech. only or B.Arch/B.Planning only is Rs.300/- and Rs.150/- for General and SC/ST candidates respectively. Candidates appearing for both BE/B.Tech and B.Arch/B.Planning together should send their application form along with additional fee in the form of Demand Draft of Rs.200/- for General and Rs.100/- for SC/ST candidates in favour of Secretary, CBSE, payable at Delhi/New Delhi.
To obtain Information Bulletin containing Application Form by post, candidates should send their request to The Joint Secretary (AIEEE), Central Board of Secondary Education, 17, Rouse Avenue, Institutional Area (Near Bal Bhavan), New Delhi-110002, along with a bank draft of Rs.350/- for General Category and Rs.200/- for SC/ST candidates in favour of The Secretary, CBSE, payable at Delhi/New Delhi and a self addressed envelope of 12”×10”.

Important information at a glance:

(1) a. Sale of AIEEE Information Bulletin containing Application Form - 01.12.2006 to 05.01. 2007

b. Online submission of application on website http://www.aieee.nic.in/ - 01.12.2006 to 05.01.2007

(2) Last date for
a. Receipt of request for Information Bulletin and Application Form by Post at CBSE Office, Shiksha Sadan, 17, Rouse Avenue, Institutional Area, (Near Bal Bhawan), New Delhi-110002 - 25.12.2006

b. Sale of Information Bulletin at designated branches of Syndicate Bank, Regional Offices of the CBSE and designated institutions - 05.01.2007

c. Online submission of applications - 05.01.2007

d. Receipt of complete applications “by post” including Registration Forms with Bank Draft at AIEEE Unit, CBSE, Shiksha Sadan, 17, Rouse Avenue, Institutional Area, (Near Bal Bhawan), New Delhi – 110002 -
10.01.2007

(3) Date of dispatch of Admit Card - 09.03.2007 to 26.03.2007
(4) Issue/dispatch of duplicate admit card(on request only with fee of Rs. 50/- + postal charges of Rs. 30/- extra for out station candidate): - 10.04.2007 to 29.04.2007 (By Hand)
- 10.04.2007 to 20.04.2007 (By Post)

(5) Dates of Examination

PAPER – 1: 29.04.2007 (0930-1230 hrs)

PAPER – 2: 29.04.2007 (1400-1700 hrs)
Visit the site www.aieee.nic.in for complete information including details of ‘online’ application, extra information for candidates opting for examination centres in foreign countries, syllabus etc.

Questions involving motion in a lift

You may often encounter questions involving motion in a lift. Generally such questions are simple, but you should have no confusion in this context. So keep the following points in mind:
The weight of a body of mass ‘m’ in a lift can be remembered as m(g-a) in all situations if you apply the proper sign to the acceleration ‘a’ of the lift. The acceleration due to gravity ‘g’ always acts vertically downwards and its sign is taken as positive. The following cases can arise in this connection:
(1) Lift moving down with acceleration of magnitude ‘a’:
In this case ‘a’ also is positive and the weight is m(g-a) which is less than the real weight of the body (when it is at rest).
(2) Lift moving up with acceleration:
In this case ‘a’ is negative and the weight is m[g-(-a)] = m(g+a).
(3) Lift moving down with retardation (going to stop while moving down):
In this case also ‘a’ is negative and the weight is m[g-(-a)] = m(g+a) which is greater than the actual weight.
(4) Lift moving up with retardation (going to stop while moving up):
In this case ‘a’ is positive and the weight is m(g-a)
(5) Lift moving up or down with uniform velocity:
In this case ‘a’ is zero and the weight is mg.
(6) Lift moving down with acceleration of magnitude ‘g’ (falling freely under gravity as is the case when the rope carrying the lift breaks):
In this case ‘a’ is positive and the weight is m(g-g) which is zero.
Let us now consider the following question:

A light, frictionless pulley is suspended from the ceiling using a light string. Another light string is passed over the pulley. If unequal masses M₁ and M₂ (M₁ >M₂) are attached to the ends of this string and the system is released, what will be tension ‘T’ in the string that carries the pulley?
(a) (M₁+M₂)g (b) (M₁- M₂)g (c) M₁M₂g/(M₁+M₂) (d) 2M₁M₂g/(M₁+M₂) (e) 4M₁M₂g/(M₁+M₂)

The tension in the string that caries the pulley is due to the weights of the masses m1 and m2. But these weights are not M1g and M2g but M₁(g-a) and M₂(g+a) because M₁ is moving down with acceleration ‘a’ and M2 is moving up with acceleration ‘a’. The magnitude of ‘a’ is given by
a = Net driving force/ Total mass moved = (M1- M2)g/(M1+M2)
Therefore, tension, T = M₁(g-a) + M₂(g+a) = 4M₁M₂g/(M₁+M₂) on substituting for ‘a’. The correct option is (e)
Now consider the following simple question:
A man of mass 70 kg records his weight on a weighing scale placed inside a lift. The ratio of the weight of the man recorded when the lift is ascending with a uniform speed of 1m/s to the weight recorded when the lift is descending with a uniform speed of 2m/s will be
(a) 2 (b) 0.5 (c) 3 (d) 1.5 (e) 1
You should note that his weight is the same as his real weight (when he is at rest). There is no distinction between state of rest and uniform motion. To put this in a different manner, the acceleration of the lift is zero and hence his weight is unchanged. The ratio of the weights is therefore equal to unity [Option (e)].
Now consider the following MCQ:
What is the minimum acceleration with which a fire man can slide down a rope whose breaking strength is 80% of his weight?
(a) 0.2g (b) 0.1g (c) ) 0.9g (d) g (e) 1.1g
This is a situation where the fire man has a rope of insufficient strength. He wants to come down from an unsafe height. He can use the rope to slide down with acceleration so that his weight is reduced (as in a lift moving down with acceleration). His minimum acceleration ‘a’ is given by
m(g-a) = 0.8mg
We have equated his reduced weight to the breaking strength of the rope. The value of ‘a’ is 0.2g [Option (a)]. Note that the rope will break if the acceleration of the fire man is less than this and also that he can use a weaker rope if his acceleration is greater than this.

Multiple Choice Questions on Matter Waves

The following MCQ on de Broglie waves appeared in Kerala Medical Entrance 2004 test paper:
An electron and a proton have the same de Broglie wave length. Then the kinetic energy of the electron is
(a) zero (b) infinity (c) equal to the kinetic energy of the proton (d) greater than the kinetic energy of the proton (e) none of these
Since the de Broglie wave length λ is given by λ = h/p where ‘h’ is Planck’s constant and ‘p’ is the linear momentum, we have p = h/λ. Therefore, the proton and the electron given in the question have the same linear momentum. The kinetic energy (E) is given by E = p²/2m. Since the mass (m) of the electron is less than that of the proton, it follows that the kinetic energy of the electron is greater [Option (d)].
Now consider the following MCQ:
A nucleus of mass ‘M’ at rest emits an α-particle of mass ‘m’. The de Broglie wave lengths of the α-particle and residual nucleus will be in the ratio
(a) m : M (b) (M+m) : m (c) M : m (d) √m:√M (e) 1 : 1
After the α-emission, the α-particle and the residual nucleus will fly off in opposite directions in accordance with the law of conservation of linear momentum. Since the momenta are equal in magnitude, the de Broglie wave lengths are the same and the ratio is 1:1 [Option (e)].
The following MCQ tests your basic knowledge regarding the electron orbits in the hydrogen atom:
The ratio of the de Broglie wave lengths of the electron in the first and the third orbits in the hydrogen atom is
(a) 1 : 1 (b) 1 : 3 (c) 1 : 9 (d) 1 : 6 (e) 1 : 27
You should remember that the orbit of quantum number ‘n’ is made of ‘n’ complete waves so that we have generally 2πr_n = nλ_n where r_nn is the radius of the nth orbit and λ_nn is the wave length of the electron in the n^th orbit. So, we have 2πr₁ = λ₁ for the first orbit and 2πr₃ = 3×λ₃ for the third orbit. Therefore λ₁/λ₃ = 3r₁/ r₃.
But the orbital radius ‘r’ is directly proportional to n². Therefore, λ₁/λ₃ = 3/9 = ⅓. The correct option is (b).
Now consider the following MCQ:
The kinetic energy of an electron is the same as that of a photon of wave length 3100 A.U. What is the wave length of this electron?
(a) 4 A.U. (b) 5.4 A.U. (c) 6.1 A.U. (d) 7.6 A.U. (e) 12.4 A.U.
In the case of a photon, the product λE = 12400 where the wave length λ is in Angstrom Unit (A.U.) and the energy E is in electron volt. Therefore, the energy of the photon of wave length 3100 A.U. is 12400/3100 = 4eV. Since the kinetic energy of the electron is the same as that of the photon, it follows that the kinetic energy of the electron is 4eV, which means this electron was accelerated by 4 volts. The wave length of an electron accelerated by ‘V’ volt is √(150/V) A.U. The answer to the problem is thus √(150/4) A.U. = √(37.5) = 6.1 A.U. approximately [Option (c)].
[You should remember that you can use the above simple relation for the wave length of an electron at small accelerating voltages only (in other words, at non-relativistic speeds only)].

Questions on Communication Systems

Let us discuss the following multiple choice question which appeared in the Kerala Engineering entrance 2005 test paper:
If a radio receiver is tuned to 855kHz radio wave, the frequency of local oscillator in kHz is
(a) 1510 (b) 455 (c) 1310 (d) 1500 (e) 855

Since the radio frequency mentioned in the question is 855 kHz, it must an AM (amplitude modulation) receiver. AM receivers are super heterodyne receivers with intermediate frequency (IF) equal to 455 kHz, by convention.. So the local oscillator in the receiver has a frequency higher by 455 kHz compared to the frequency of the received radio wave. Its frequency is therefore (855+455) kHz = 1310 kHz.
Consider the following MCQ which high lights the component frequencies in an AM wave:
A carrier of frequency 1.2 MHz is amplitude modulated by a microphone output of frequency 800 Hz. The frequencies present in the amplitude modulated carrier are (in kHz)
(a)1.2 MHz and 800 Hz (b) 400 kHz, 1.2 MHz and 2 MHz (c)1.04 MHz,1.2 MHz and 2 MHz (d) 1.1992 MHz, 1.2 MHz and 1.2008 MHz (e) 1.1992 MHz and 1.2008 MHz
The correct option is (d) since the AM wave contains the carrier frequency, the upper side frequency (which is the sum of the carrier frequency and the modulating signal frequency) and the lower side frequency (which is the difference between the carrier frequency and the modulating signal frequency). While finding the upper and lower side frequencies, it will be convenient if you convert the frequencies in to kilohertz or megahertz. Thus, the lower side frequency = (1200 – 0.8)kHz = 1199.2 kHz = 1.1992 MHz. Similarly, the upper side frequency = (1200 + 0.8) kHz = 1200.8 kHz = 1.2008 MHz.
Now consider the following question:
A semiconductor is to be selected for use as a photo detector to detect radiation of wavelength 620 nm. The appropriate semiconductor will have a band gap of
(a) 2eV (b) 2.2eV (c) 2.4eV (d) 2.8eV (e) both (a) and (b)
The band gap of the semiconductor should not exceed the energy of the photon. The wavelength of the photon being 620 nm which is equal to 6200 Angstrom Units (A.U), its energy is 12400/6200 = 2.0 eV (Note that the product of the wave length in A. U. and the energy in eV for a photon is 12400). The option (a) alone is therefore suitable.
The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:
If α and β are the current gains in the CB and CE configurations respectively of the transistor circuit, then (β – α)/αβ =
(a) ∞ (b) 1 (c) 2 (d) 0.5 (e) zero
Since β = α/(1- α), the given ratio, (β – α)/αβ =[α/(1- α) – α]/αβ = [1/(1- α) -1]/β =[1-(1- α)]/(1-α)β = α/(1- α)β = β/β =1. So, the correct option is (b).